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

Optimal. Leaf size=197 \[ \frac {2 \sqrt {2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {4 (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac {8 a (8 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \]

[Out]

2*a^(3/2)*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-8/35*a*(7*I*A+8*B)*(a+I*a*ta
n(d*x+c))^(1/2)/d+2/35*a*(7*I*A+8*B)*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d+2/7*I*a*B*(a+I*a*tan(d*x+c))^(1/2
)*tan(d*x+c)^3/d-4/105*(21*I*A+19*B)*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]  time = 0.53, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3594, 3597, 3592, 3527, 3480, 206} \[ \frac {2 \sqrt {2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {4 (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac {8 a (8 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*a^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a*((7*I)*A + 8*B)*S
qrt[a + I*a*Tan[c + d*x]])/(35*d) + (2*a*((7*I)*A + 8*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) + (
((2*I)/7)*a*B*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (4*((21*I)*A + 19*B)*(a + I*a*Tan[c + d*x])^(3/2)
)/(105*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3594

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

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]

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {2}{7} \int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (7 A-6 i B)+\frac {1}{2} a (7 i A+8 B) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {4 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-a^2 (7 i A+8 B)+\frac {1}{2} a^2 (21 A-19 i B) \tan (c+d x)\right ) \, dx}{35 a}\\ &=\frac {2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}+\frac {4 \int \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{2} a^2 (21 A-19 i B)-a^2 (7 i A+8 B) \tan (c+d x)\right ) \, dx}{35 a}\\ &=-\frac {8 a (7 i A+8 B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}-(2 a (A-i B)) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {8 a (7 i A+8 B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}+\frac {\left (4 a^2 (i A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {2 \sqrt {2} a^{3/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 a (7 i A+8 B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d}\\ \end {align*}

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Mathematica [A]  time = 4.45, size = 239, normalized size = 1.21 \[ \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac {2 \sqrt {2} (B+i A) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}-\frac {1}{210} (\tan (c+d x)+i) \sec ^{\frac {5}{2}}(c+d x) (21 (17 A-18 i B) \cos (c+d x)+(147 A-158 i B) \cos (3 (c+d x))+42 i A \sin (c+d x)+42 i A \sin (3 (c+d x))-7 B \sin (c+d x)+53 B \sin (3 (c+d x)))\right )}{d \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*((2*Sqrt[2]*(I*A + B)*ArcSinh[E^(I*(c + d*x))])/((E^(I*(c +
d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) - (Sec[c + d*x]^(5/2)*(21*(17*A - (18
*I)*B)*Cos[c + d*x] + (147*A - (158*I)*B)*Cos[3*(c + d*x)] + (42*I)*A*Sin[c + d*x] - 7*B*Sin[c + d*x] + (42*I)
*A*Sin[3*(c + d*x)] + 53*B*Sin[3*(c + d*x)])*(I + Tan[c + d*x]))/210))/(d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] +
B*Sin[c + d*x]))

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fricas [B]  time = 0.56, size = 471, normalized size = 2.39 \[ \frac {105 \, \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (8 i \, A + 8 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) - 105 \, \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (8 i \, A + 8 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) + \sqrt {2} {\left ({\left (-1512 i \, A - 1688 \, B\right )} a e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (-3192 i \, A - 2968 \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (-2520 i \, A - 3080 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (-840 i \, A - 840 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{420 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(105*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*
e^(2*I*d*x + 2*I*c) + d)*log(((8*I*A + 8*B)*a^2*e^(I*d*x + I*c) + sqrt(2)*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a
^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) - 1
05*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d
*x + 2*I*c) + d)*log(((8*I*A + 8*B)*a^2*e^(I*d*x + I*c) - sqrt(2)*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*
(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) + sqrt(2)*(
(-1512*I*A - 1688*B)*a*e^(7*I*d*x + 7*I*c) + (-3192*I*A - 2968*B)*a*e^(5*I*d*x + 5*I*c) + (-2520*I*A - 3080*B)
*a*e^(3*I*d*x + 3*I*c) + (-840*I*A - 840*B)*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(6*I*d*
x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + 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)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.29, size = 164, normalized size = 0.83 \[ -\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5}+\frac {A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5}-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}}{3}-i a^{3} B \sqrt {a +i a \tan \left (d x +c \right )}+A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}-a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 153, normalized size = 0.78 \[ -\frac {i \, {\left (105 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 30 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} B a + 42 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (A + i \, B\right )} a^{2} - 70 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a^{3} + 210 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{4}\right )}}{105 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*I*(105*sqrt(2)*(A - I*B)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) +
sqrt(I*a*tan(d*x + c) + a))) - 30*I*(I*a*tan(d*x + c) + a)^(7/2)*B*a + 42*(I*a*tan(d*x + c) + a)^(5/2)*(A + I
*B)*a^2 - 70*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a^3 + 210*sqrt(I*a*tan(d*x + c) + a)*(A - I*B)*a^4)/(a^3*d)

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mupad [B]  time = 7.42, size = 211, normalized size = 1.07 \[ -\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}-\frac {A\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}-\frac {2\,B\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a\,d}+\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a\,d}-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a^2\,d}-\frac {\sqrt {2}\,A\,{\left (-a\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,2{}\mathrm {i}}{d}-\frac {\sqrt {2}\,B\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x))*tan(c + d*x)**2, x)

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