3.155 \(\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=152 \[ -\frac {(-1)^{3/4} \sqrt {a} (2 A-i B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

-(-1)^(3/4)*(2*A-I*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)/d-(1+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))*a^(1/2)/d+B*tan(d*x+c)^(1/2)*(a+I*a*tan(
d*x+c))^(1/2)/d

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Rubi [A]  time = 0.49, antiderivative size = 152, 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 = {3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac {(-1)^{3/4} \sqrt {a} (2 A-i B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

-(((-1)^(3/4)*Sqrt[a]*(2*A - I*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/
d) - ((1 + I)*Sqrt[a]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d +
(B*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d

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 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 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 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {a B}{2}+\frac {1}{2} a (2 A-i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a}\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+(-i A-B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\frac {(2 i A+B) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\left (2 a^2 (A-i B)\right ) \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 {(a (2 i A+B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {(1+i) \sqrt {a} (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 )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {(a (2 i A+B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {(1+i) \sqrt {a} (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 )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {(a (2 i A+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 {\sqrt [4]{-1} \sqrt {a} (2 i A+B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} (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 )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\\ \end {align*}

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Mathematica [B]  time = 4.73, size = 560, normalized size = 3.68 \[ -\frac {e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \sqrt {a+i a \tan (c+d x)} \left (8 (B+i A) \left (1+e^{2 i (c+d x)}\right ) \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )-i \sqrt {2} (2 A-i B) \left (1+e^{2 i (c+d x)}\right ) \log \left (-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+2 i \sqrt {2} A e^{2 i (c+d x)} \log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+2 i \sqrt {2} A \log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-8 B e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}+\sqrt {2} B e^{2 i (c+d x)} \log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+\sqrt {2} B \log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )}{4 \sqrt {2} d \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

-1/4*(Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c
+ d*x)))]*(-8*B*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))] + 8*(I*A + B)*(1 + E^((2*I)*(c + d*x)))*Log[E^(
I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] - I*Sqrt[2]*(2*A - I*B)*(1 + E^((2*I)*(c + d*x)))*Log[1 - 3*E^(
(2*I)*(c + d*x)) - 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + (2*I)*Sqrt[2]*A*Log[1 - 3*E^((2
*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*B*Log[1 - 3*E^((2*I)*(c +
d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + (2*I)*Sqrt[2]*A*E^((2*I)*(c + d*x))*Log[1
- 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*B*E^((2*I)*(c +
d*x))*Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[a + I*a
*Tan[c + d*x]])/(Sqrt[2]*d*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[Sec[c + d*x]])

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fricas [B]  time = 0.67, size = 656, normalized size = 4.32 \[ \frac {2 \, \sqrt {2} B \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}} e^{\left (i \, d x + i \, c\right )} + d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 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}} + 2 \, d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 i \, A + B}\right ) - d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 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}} - 2 \, d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 i \, A + B}\right ) - d \sqrt {\frac {{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\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}} + d \sqrt {\frac {{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + d \sqrt {\frac {{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\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}} - d \sqrt {\frac {{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(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) + d*sqrt((4*I*A^2 + 4*A*B - I*B^2)*a/d^2)*log((sqrt(2)*((2*I*A + B)*e^(2*I*d*x + 2*I*c) + 2*
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)) + 2*d*
sqrt((4*I*A^2 + 4*A*B - I*B^2)*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(2*I*A + B)) - d*sqrt((4*I*A^2 + 4*A*B
- I*B^2)*a/d^2)*log((sqrt(2)*((2*I*A + B)*e^(2*I*d*x + 2*I*c) + 2*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)) - 2*d*sqrt((4*I*A^2 + 4*A*B - I*B^2)*a/d^2)*e^(I*
d*x + I*c))*e^(-I*d*x - I*c)/(2*I*A + B)) - d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)*a/d^2)*log((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)) + d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) + d*s
qrt((2*I*A^2 + 4*A*B - 2*I*B^2)*a/d^2)*log((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)) - d*sqrt((2*I*A^2 + 4*A*B - 2*I
*B^2)*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.39, size = 713, normalized size = 4.69 \[ -\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 \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 +2 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 ) \sqrt {-i a}\, \tan \left (d x +c \right ) a +i 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 -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 ) \tan \left (d x +c \right ) a -2 i 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}-i 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}\, a +2 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}\, \tan \left (d x +c \right )+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}\, \tan \left (d x +c \right ) a +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 ) a +2 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 ) \sqrt {-i a}\, a \right )}{2 d \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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x)

[Out]

-1/2/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(I*B*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*ta
n(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+2*I*A*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*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-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))*tan(d*x+c)*a-2*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-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)*a+2*B*(
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)+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+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))*a+2*A*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)*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)+I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \sqrt {\tan \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(I*a*tan(d*x + c) + a)*sqrt(tan(d*x + c)), x)

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mupad [B]  time = 24.35, size = 2225, normalized size = 14.64 \[ \text {result too large to display} \]

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]

- ((B*tan(c + d*x)^(3/2)*2i)/(d*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^3) + (2*B*tan(c + d*x)^(1/2))/(a*d*(
(a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))))/(tan(c + d*x)^2/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^4 - 1/a^2
+ (tan(c + d*x)*2i)/(a*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - ((-a)^(1/2)*atan((A^4*(-a)^(21/2)*tan(
c + d*x)^(1/2)*(7168 - 7168i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512i - 40
96*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)
^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A^2*B^2
*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a + a*tan(
c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2
)) + (B^4*(-a)^(21/2)*tan(c + d*x)^(1/2)*(1024 - 1024i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*
3584i + B^4*a^10*512i - 4096*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x)
)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a
^(1/2))^2 + (10240*A^2*B^2*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c
+ d*x)*4096i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d
*x)*1i)^(1/2) - a^(1/2))^2)) + (A*B^3*(-a)^(21/2)*tan(c + d*x)^(1/2)*(8192 + 8192i))/(((a + a*tan(c + d*x)*1i)
^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512i - 4096*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i -
(3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a +
a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A^2*B^2*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(
1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*
x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - (A^3*B*(-a)^(21/2)*tan(c + d*x)^(1/2)*(20480 + 2048
0i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512i - 4096*A*B^3*a^10 + 10240*A^3*
B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (512
*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A^2*B^2*a^11*tan(c + d*x))/((a +
a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/
2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - (A^2*B^2*(-a)^(21/2)*
tan(c + d*x)^(1/2)*(20480 - 20480i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512
i - 4096*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*
x)*1i)^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A
^2*B^2*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a +
a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1
/2))^2)))*(A*1i + B)*(1 - 1i))/d - ((-1)^(1/4)*a^(1/2)*atan(((-1)^(1/4)*A^5*tan(c + d*x)^(1/2)*25690112i)/(a^(
15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8
- (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)) + (262144*(-1)^(1/4)*B^5*tan(c +
d*x)^(1/2))/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (367
0016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)) + ((-1)^(1/4)*A*B
^4*tan(c + d*x)^(1/2)*3670016i)/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5
*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8
)) + (56885248*(-1)^(1/4)*A^4*B*tan(c + d*x)^(1/2))/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((2569
0112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (4
8234496*A^3*B^2)/a^8)) - (19398656*(-1)^(1/4)*A^2*B^3*tan(c + d*x)^(1/2))/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(
1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B
^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)) - ((-1)^(1/4)*A^3*B^2*tan(c + d*x)^(1/2)*48234496i)/(a^(15/2)*((a
+ a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*
56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)))*(2*A - B*1i)*2i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*(A + B*tan(c + d*x))*sqrt(tan(c + d*x)), x)

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