3.180 \(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\)
Optimal. Leaf size=156 \[ \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]
[Out]
-2*(-1)^(1/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d/a^(1/2)-(1/2+1/2*I)*(A-
I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d/a^(1/2)+(I*A-B)*tan(d*x+c)^(1/2)/d/(a+
I*a*tan(d*x+c))^(1/2)
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Rubi [A] time = 0.48, antiderivative size = 156, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used =
{3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
(-2*(-1)^(1/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - ((1
/2 + I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + ((
I*A - B)*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 3544
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Rule 3595
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Rule 3599
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]
Rule 3601
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Rubi steps
\begin {align*} \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (i A-B)+i a B \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a^2}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}+\frac {B \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^2}-\frac {(i A+B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(a (A-i B)) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(2 B) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(2 B) \operatorname {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 4.00, size = 183, normalized size = 1.17 \[ \frac {\sqrt {\tan (c+d x)} \left (i (A+i B) \sqrt {-1+e^{2 i (c+d x)}}-i (A-i B) e^{i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+2 \sqrt {2} B e^{i (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{d \sqrt {-1+e^{2 i (c+d x)}} \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
((I*(A + I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))] - I*(A - I*B)*E^(I*(c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E
^((2*I)*(c + d*x))]] + 2*Sqrt[2]*B*E^(I*(c + d*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d
*x))]])*Sqrt[Tan[c + d*x]])/(d*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[a + I*a*Tan[c + d*x]])
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fricas [B] time = 0.88, size = 705, normalized size = 4.52 \[ \frac {{\left (a d \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} {\left (B e^{\left (3 i \, d x + 3 i \, c\right )} + B e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + {\left (3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) - a d \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} {\left (B e^{\left (3 i \, d x + 3 i \, c\right )} + B e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - {\left (3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) - a d \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {a d \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + a d \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {a d \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left ({\left (2 i \, A - 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - 2 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/4*(a*d*sqrt(-4*I*B^2/(a*d^2))*e^(I*d*x + I*c)*log(52/605*(4*sqrt(2)*(B*e^(3*I*d*x + 3*I*c) + B*e^(I*d*x + I*
c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + (3*a*d*e^
(2*I*d*x + 2*I*c) - a*d)*sqrt(-4*I*B^2/(a*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) - a*d*sqrt(-4*I*B^2/(a*d^2))*e^(
I*d*x + I*c)*log(52/605*(4*sqrt(2)*(B*e^(3*I*d*x + 3*I*c) + B*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - (3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-4*I*
B^2/(a*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) - a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(I*d*x + I*c)*log
((a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A
+ B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(4*I*A +
4*B)) + a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(I*d*x + I*c)*log(-(a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B
^2)/(a*d^2))*e^(I*d*x + I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(4*I*A + 4*B)) + sqrt(2)*((2*I*A - 2*B)*e^(
2*I*d*x + 2*I*c) + 2*I*A - 2*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(a*d)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\tan \left (d x + c\right )}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*sqrt(tan(d*x + c))/sqrt(I*a*tan(d*x + c) + a), x)
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maple [B] time = 0.42, size = 900, normalized size = 5.77 \[ \frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i B \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, \left (\tan ^{2}\left (d x +c \right )\right ) a +2 i A \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, \tan \left (d x +c \right ) a -A \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, \left (\tan ^{2}\left (d x +c \right )\right ) a -8 i B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right ) a -i B \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a +4 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right ) a +2 B \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a -4 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+A \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +4 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-4 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +4 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x)
[Out]
1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a*(I*B*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*2^(1/2)*tan(d*x+c)^2*a+2*I*A*ln(-(-2*2^(1/2)*(-I
*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*2^(1/2)*tan(d*
x+c)*a-A*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I)
)*(I*a)^(1/2)*2^(1/2)*tan(d*x+c)^2*a-8*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I
*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)*a-I*B*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*
x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*2^(1/2)*a+4*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+4*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*
(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a+2*B*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(
a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a-4*I*A*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+A*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a+4*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^
(1/2)*(-I*a)^(1/2)*tan(d*x+c)-4*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2
)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a+4*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2))/(a*tan(d*x
+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)^2
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.
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mupad [B] time = 24.21, size = 4040, normalized size = 25.90 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((tan(c + d*x)^(1/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(1/2),x)
[Out]
-(A*a^(5/2)*tan(c + d*x)^(1/2)*4i + 4*A*a^(5/2)*tan(c + d*x)^(3/2) - 4*B*a^(5/2)*tan(c + d*x)^(1/2) + B*a^(5/2
)*tan(c + d*x)^(3/2)*4i + A*a^(3/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i - 4*B*a^(3/2)*tan(c + d*x)^(
1/2)*(a + a*tan(c + d*x)*1i) + (1i/8)^(1/2)*A*(-a)^(5/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)
^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(3
1/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3
/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*
x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*8i + 8*(1i/8)^(1/2)*B*(-a)^(5
/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(
3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)
*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^
(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a
^16 + a^16*tan(c + d*x)^2)) - A*a^2*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i - 4*A*a^2*tan(c + d*x)
^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) + 8*B*a^2*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2) - B*a^2*tan(c
+ d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2)*4i + 4i^(1/2)*B*(-a)^(5/2)*atan((4i^(1/2)*a^12*tan(c + d*x)^(1/2))/
(2*(-a)^(23/2)*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i)^(1/2)))*8i + (1i/8)^(1/2)*A*(-a)^(5/2)*atanh(((
1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*
tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/
8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a
*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*t
an(c + d*x)^2))*tan(c + d*x)^2*8i + 8*(1i/8)^(1/2)*B*(-a)^(5/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c
+ d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*
(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c +
d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan
(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*tan(c + d*x)^2 - (1i/8
)^(1/2)*A*(-a)^(3/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/
2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c
+ d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15
/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d
*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*(a + a*tan(c + d*x)*1i)*8i + 4i^(1/2)*B*(-a)^(5/2)*tan(c + d*x)^2
*atan((4i^(1/2)*a^12*tan(c + d*x)^(1/2))/(2*(-a)^(23/2)*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i)^(1/2))
)*8i - 8*(1i/8)^(1/2)*B*(-a)^(3/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(
1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2
)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a
)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a
+ a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*(a + a*tan(c + d*x)*1i) - 4i^(1/2)*B*(-a)^(3/2)*ata
n((4i^(1/2)*a^12*tan(c + d*x)^(1/2))/(2*(-a)^(23/2)*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i)^(1/2)))*(a
+ a*tan(c + d*x)*1i)*8i - (1i/8)^(1/2)*A*(-a)^(1/2)*a^(3/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c +
d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a
)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x
)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c
+ d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*(a + a*tan(c + d*x)*1i)^(
1/2)*12i + (1i/8)^(1/2)*A*(-a)^(3/2)*a^(1/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i -
4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c +
d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)
^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a
^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*(a + a*tan(c + d*x)*1i)^(1/2)*4i - 12*(1i
/8)^(1/2)*B*(-a)^(1/2)*a^(3/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)
*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a
+ a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(1
5/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a
*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*(a + a*tan(c + d*x)*1i)^(1/2) + 4*(1i/8)^(1/2)*B*(-a)^(
3/2)*a^(1/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c
+ d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*
1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(
c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^
(1/2) + a^16 + a^16*tan(c + d*x)^2))*(a + a*tan(c + d*x)*1i)^(1/2) - 4i^(1/2)*B*(-a)^(1/2)*a^(3/2)*atan((4i^(1
/2)*a^12*tan(c + d*x)^(1/2))/(2*(-a)^(23/2)*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i)^(1/2)))*(a + a*tan
(c + d*x)*1i)^(1/2)*12i + 4i^(1/2)*B*(-a)^(3/2)*a^(1/2)*atan((4i^(1/2)*a^12*tan(c + d*x)^(1/2))/(2*(-a)^(23/2)
*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i)^(1/2)))*(a + a*tan(c + d*x)*1i)^(1/2)*4i + 4*(1i/8)^(1/2)*A*(
-a)^(1/2)*a^(3/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*
tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c +
d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)
*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)
*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2))*tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/2) + 4*(1i/8)^(1/2)*A*(-a)^(
3/2)*a^(1/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c
+ d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*
1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(
c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^
(1/2) + a^16 + a^16*tan(c + d*x)^2))*tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*B*(-a)^(1/2)*a^
(3/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)
^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/
2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x
)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) +
a^16 + a^16*tan(c + d*x)^2))*tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/2)*4i - (1i/8)^(1/2)*B*(-a)^(3/2)*a^(1/2
)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/
2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8
i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1
/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^1
6 + a^16*tan(c + d*x)^2))*tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/2)*4i + 4*4i^(1/2)*B*(-a)^(1/2)*a^(3/2)*tan(
c + d*x)*atan((4i^(1/2)*a^12*tan(c + d*x)^(1/2))/(2*(-a)^(23/2)*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i
)^(1/2)))*(a + a*tan(c + d*x)*1i)^(1/2) + 4*4i^(1/2)*B*(-a)^(3/2)*a^(1/2)*tan(c + d*x)*atan((4i^(1/2)*a^12*tan
(c + d*x)^(1/2))/(2*(-a)^(23/2)*a^(1/2) - 2*(-a)^(23/2)*(a + a*tan(c + d*x)*1i)^(1/2)))*(a + a*tan(c + d*x)*1i
)^(1/2))/(4*a^3*d + 4*a^3*d*tan(c + d*x)^2 + 4*a^2*d*(a + a*tan(c + d*x)*1i) - 8*a^(5/2)*d*(a + a*tan(c + d*x)
*1i)^(1/2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2),x)
[Out]
Integral((A + B*tan(c + d*x))*sqrt(tan(c + d*x))/sqrt(I*a*(tan(c + d*x) - I)), x)
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