3.183 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=191 \[ -\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {(-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (B+i A) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

(1/2+1/2*I)*(I*A+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)+1/3*(7*I*A-9*B)
*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(1/2)+(A+I*B)/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2)-1/3*(5*A+3*
I*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(3/2)

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Rubi [A]  time = 0.55, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ -\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {(-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (B+i A) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((1/2 + I/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) +
(A + I*B)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) - ((5*A + (3*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3
*a*d*Tan[c + d*x]^(3/2)) + (((7*I)*A - 9*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d*Sqrt[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*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] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (5 A+3 i B)-2 a (i A-B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (7 i A-9 B)-\frac {1}{2} a^2 (5 A+3 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^3}\\ &=\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(7 i A-9 B) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {3 a^3 (A-i B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(7 i A-9 B) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {(A-i B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(7 i A-9 B) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {(a (i A+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 {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+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 {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(7 i A-9 B) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 4.31, size = 221, normalized size = 1.16 \[ \frac {e^{-i (c+d x)} (A+B \tan (c+d x)) \left (3 (B+i A) e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+i A \left (-18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3\right )-3 B \left (-6 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+1\right )\right )}{6 d \left (-1+e^{2 i (c+d x)}\right ) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-3*B*(1 - 6*E^((2*I)*(c + d*x)) + 5*E^((4*I)*(c + d*x))) + I*A*(3 - 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c +
d*x))) + 3*(I*A + B)*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2
*I)*(c + d*x))]])*(A + B*Tan[c + d*x]))/(6*d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*(A*Cos[c + d*x] + B*Si
n[c + d*x])*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.73, size = 518, normalized size = 2.71 \[ -\frac {\sqrt {2} {\left (2 \, {\left (7 \, A + 15 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (11 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 30 \, {\left (A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, A + 6 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 3 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} \log \left (\frac {i \, a d \sqrt {\frac {-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 3 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} \log \left (\frac {-i \, a d \sqrt {\frac {-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right )}{12 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(sqrt(2)*(2*(7*A + 15*I*B)*e^(6*I*d*x + 6*I*c) - 2*(11*A + 3*I*B)*e^(4*I*d*x + 4*I*c) - 30*(A + I*B)*e^(
2*I*d*x + 2*I*c) + 6*A + 6*I*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)) + 3*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt((-2*I*A^
2 - 4*A*B + 2*I*B^2)/(a*d^2))*log((I*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)) - 3*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I
*d*x + I*c))*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)/(a*d^2))*log((-I*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*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I
*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/(sqrt(I*a*tan(d*x + c) + a)*tan(d*x + c)^(5/2)), x)

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maple [B]  time = 0.36, size = 746, normalized size = 3.91 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-3 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 ) \left (\tan ^{2}\left (d x +c \right )\right ) a -36 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )-6 i B \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 ) \left (\tan ^{3}\left (d x +c \right )\right ) a +3 B \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 ) \left (\tan ^{4}\left (d x +c \right )\right ) a +36 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+28 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )+60 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+6 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 ) \left (\tan ^{3}\left (d x +c \right )\right ) a +3 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 ) \left (\tan ^{4}\left (d x +c \right )\right ) a -3 B \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 ) \left (\tan ^{2}\left (d x +c \right )\right ) a +24 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )+8 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{12 d a \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \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((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x)

[Out]

1/12/d*(a*(1+I*tan(d*x+c)))^(1/2)/a/tan(d*x+c)^(3/2)*(-3*I*A*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)^2*a-36*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(
1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-6*I*B*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)^3*a+3*B*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)^4*a+36*A*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+28*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+60
*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+6*A*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)^3*a+3*I*A*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)^4*a
-3*B*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)^2*a+24*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+8*A*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(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((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))/(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**(5/2)), x)

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