3.194 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=240 \[ \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(1/8+1/8*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^(5/2)/d+1/60*(151*A+41*
I*B)/a^2/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-1/60*(317*A+67*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/tan(d*
x+c)^(1/2)+1/5*(A+I*B)/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2)+1/30*(17*A+7*I*B)/a/d/tan(d*x+c)^(1/2)/(a+I
*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.80, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((1/8 + I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) +
(A + I*B)/(5*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + (17*A + (7*I)*B)/(30*a*d*Sqrt[Tan[c + d*x]]
*(a + I*a*Tan[c + d*x])^(3/2)) + (151*A + (41*I)*B)/(60*a^2*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) -
((317*A + (67*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*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 {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a (11 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 (83 A+13 i B)-a^2 (17 i A-7 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (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}{8} a^3 (317 A+67 i B)-\frac {1}{4} a^3 (151 i A-41 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {2 \int \frac {15 a^4 (i A+B) \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{15 a^7}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {(i A+B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {(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 )}{4 a d}\\ &=\frac {\left (\frac {1}{8}+\frac {i}{8}\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 )}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 10.73, size = 288, normalized size = 1.20 \[ \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x)) \left (\frac {\sqrt {2} (B+i A) e^{3 i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+\frac {2 ((19 B-149 i A) \tan (c+d x)+\cos (2 (c+d x)) ((86 B-466 i A) \tan (c+d x)-20 (23 A+4 i B))+340 A+80 i B)}{15 \sqrt {\tan (c+d x)} \sqrt {\sec (c+d x)}}\right )}{8 d (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

(Sec[c + d*x]^(3/2)*(A + B*Tan[c + d*x])*((Sqrt[2]*(I*A + B)*E^((3*I)*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c
+ d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(Sqrt[-1 + E^((2
*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]) + (2*(340*A + (80*I)*B + ((-149*I)*A + 19*B)*
Tan[c + d*x] + Cos[2*(c + d*x)]*(-20*(23*A + (4*I)*B) + ((-466*I)*A + 86*B)*Tan[c + d*x])))/(15*Sqrt[Sec[c + d
*x]]*Sqrt[Tan[c + d*x]])))/(8*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2))

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fricas [B]  time = 0.60, size = 528, normalized size = 2.20 \[ \frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} 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 ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} 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 (-463 i \, A + 83 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-269 i \, A + 19 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (220 i \, A - 80 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (29 i \, A - 19 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, 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}}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*
d^2))*log((2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*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)) - 15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((I*A^2
+ 2*A*B - I*B^2)/(a^5*d^2))*log(-(2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*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)*((-463*I*A + 83*B)*e^(8*I*d*x + 8*I*c) + (-269*I
*A + 19*B)*e^(6*I*d*x + 6*I*c) + (220*I*A - 80*B)*e^(4*I*d*x + 4*I*c) + (29*I*A - 19*B)*e^(2*I*d*x + 2*I*c) +
3*I*A - 3*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)))/(
a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/((I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(3/2)), x)

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maple [B]  time = 0.29, size = 1158, normalized size = 4.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(-4468*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+
c)^3+1268*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4+15*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))*2^(1/2)*tan(d*x+c)*a-15*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)^5*a+908*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+15*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)^5*a-60*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))*2^(1/2)*t
an(d*x+c)^2*a+60*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-5660*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+2940
*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+60*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))*2^(1/2)*tan(d*x+c)^4*a+90*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
+268*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4-90*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))*2^(1/2)*tan(d*x+c)^3*a-60*B*2^(1/2)*l
n(-(-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-1060*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-15*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)*a-420*B*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+480*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1
/2))/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)^4

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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))/tan(d*x+c)^(3/2)/(a+I*a*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 = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)), x)

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sympy [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((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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