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

Optimal. Leaf size=198 \[ \frac {8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {8 a^3 (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {8 a^3 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

[Out]

8*(-1)^(1/4)*a^3*(A-I*B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+8*a^3*(A-I*B)*tan(d*x+c)^(1/2)/d+8/3*a^3*(I*A+B
)*tan(d*x+c)^(3/2)/d-16/315*a^3*(18*A-19*I*B)*tan(d*x+c)^(5/2)/d+2/9*I*a*B*tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))
^2/d-2/63*(9*A-13*I*B)*tan(d*x+c)^(5/2)*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]  time = 0.47, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3594, 3592, 3528, 3533, 205} \[ -\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {8 a^3 (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {8 a^3 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(A - I*B)*Sqrt[Tan[c + d*x]])/d
+ (8*a^3*(I*A + B)*Tan[c + d*x]^(3/2))/(3*d) - (16*a^3*(18*A - (19*I)*B)*Tan[c + d*x]^(5/2))/(315*d) + (((2*I)
/9)*a*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2)/d - (2*(9*A - (13*I)*B)*Tan[c + d*x]^(5/2)*(a^3 + I*a^3*T
an[c + d*x]))/(63*d)

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 3528

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 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]

Rubi steps

\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}+\frac {2}{9} \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2 \left (\frac {1}{2} a (9 A-5 i B)+\frac {1}{2} a (9 i A+13 B) \tan (c+d x)\right ) \, dx\\ &=\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {4}{63} \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) \left (a^2 (27 A-25 i B)+2 a^2 (18 i A+19 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {4}{63} \int \tan ^{\frac {3}{2}}(c+d x) \left (63 a^3 (A-i B)+63 a^3 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {8 a^3 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {4}{63} \int \sqrt {\tan (c+d x)} \left (-63 a^3 (i A+B)+63 a^3 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac {8 a^3 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {8 a^3 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {4}{63} \int \frac {-63 a^3 (A-i B)-63 a^3 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {8 a^3 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {8 a^3 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {\left (504 a^6 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-63 a^3 (A-i B)+63 a^3 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {8 a^3 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {16 a^3 (18 A-19 i B) \tan ^{\frac {5}{2}}(c+d x)}{315 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {2 (9 A-13 i B) \tan ^{\frac {5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}\\ \end {align*}

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Mathematica [B]  time = 10.41, size = 496, normalized size = 2.51 \[ \frac {\cos ^4(c+d x) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\sec (c) \left (\frac {2}{7} \cos (3 c)-\frac {2}{7} i \sin (3 c)\right ) \sec ^3(c+d x) (-3 B \sin (d x)-i A \sin (d x))+\sec (c) \left (-\frac {2}{315} \cos (3 c)+\frac {2}{315} i \sin (3 c)\right ) \sec ^2(c+d x) (45 i A \sin (c)+189 A \cos (c)+135 B \sin (c)-322 i B \cos (c))+\sec (c) \left (\frac {2}{21} \cos (3 c)-\frac {2}{21} i \sin (3 c)\right ) \sec (c+d x) (37 B \sin (d x)+31 i A \sin (d x))+\sec (c) \left (\frac {2}{315} \cos (3 c)-\frac {2}{315} i \sin (3 c)\right ) (465 i A \sin (c)+1449 A \cos (c)+555 B \sin (c)-1547 i B \cos (c))+\left (-\frac {2}{9} B \sin (3 c)-\frac {2}{9} i B \cos (3 c)\right ) \sec ^4(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}-\frac {8 e^{-3 i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c
+ d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*E^((3*I
)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*
Sin[c + d*x])) + (Cos[c + d*x]^4*(Sec[c]*(1449*A*Cos[c] - (1547*I)*B*Cos[c] + (465*I)*A*Sin[c] + 555*B*Sin[c])
*((2*Cos[3*c])/315 - ((2*I)/315)*Sin[3*c]) + Sec[c]*Sec[c + d*x]^2*(189*A*Cos[c] - (322*I)*B*Cos[c] + (45*I)*A
*Sin[c] + 135*B*Sin[c])*((-2*Cos[3*c])/315 + ((2*I)/315)*Sin[3*c]) + Sec[c + d*x]^4*(((-2*I)/9)*B*Cos[3*c] - (
2*B*Sin[3*c])/9) + Sec[c]*Sec[c + d*x]^3*((2*Cos[3*c])/7 - ((2*I)/7)*Sin[3*c])*((-I)*A*Sin[d*x] - 3*B*Sin[d*x]
) + Sec[c]*Sec[c + d*x]*((2*Cos[3*c])/21 - ((2*I)/21)*Sin[3*c])*((31*I)*A*Sin[d*x] + 37*B*Sin[d*x]))*Sqrt[Tan[
c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[
c + d*x]))

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fricas [B]  time = 0.86, size = 557, normalized size = 2.81 \[ -\frac {315 \, \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 315 \, \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 16 \, {\left ({\left (957 \, A - 1051 i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 5 \, {\left (579 \, A - 547 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, {\left (171 \, A - 173 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, {\left (429 \, A - 433 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (123 \, A - 124 i \, B\right )} a^{3}\right )} \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}}}{1260 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/1260*(315*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) +
6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I
*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*
d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3)) - 315*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/
d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)
*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x +
2*I*c) - I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B
)*a^3)) - 16*((957*A - 1051*I*B)*a^3*e^(8*I*d*x + 8*I*c) + 5*(579*A - 547*I*B)*a^3*e^(6*I*d*x + 6*I*c) + 21*(1
71*A - 173*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 5*(429*A - 433*I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(123*A - 124*I*B)*a^
3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I
*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.11, size = 610, normalized size = 3.08 \[ \frac {8 i a^{3} B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}-\frac {i a^{3} A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}-\frac {6 a^{3} B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7 d}+\frac {8 i a^{3} A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}-\frac {6 a^{3} A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {i a^{3} B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}+\frac {8 a^{3} B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}-\frac {8 i a^{3} B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {8 a^{3} A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {2 i a^{3} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {2 i a^{3} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {2 i a^{3} A \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7 d}-\frac {2 a^{3} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {2 a^{3} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{3} A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}-\frac {2 i a^{3} B \left (\tan ^{\frac {9}{2}}\left (d x +c \right )\right )}{9 d}-\frac {2 i a^{3} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {2 i a^{3} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{3} B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}-\frac {2 a^{3} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {2 a^{3} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/5*I/d*a^3*B*tan(d*x+c)^(5/2)-I/d*a^3*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x
+c)^(1/2)+tan(d*x+c)))-6/7/d*a^3*B*tan(d*x+c)^(7/2)+8/3*I/d*a^3*A*tan(d*x+c)^(3/2)-6/5/d*a^3*A*tan(d*x+c)^(5/2
)+I/d*a^3*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+8/3/d*
a^3*B*tan(d*x+c)^(3/2)-8*I/d*a^3*B*tan(d*x+c)^(1/2)+8*a^3*A*tan(d*x+c)^(1/2)/d-2*I/d*a^3*A*2^(1/2)*arctan(-1+2
^(1/2)*tan(d*x+c)^(1/2))+2*I/d*a^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-2/7*I/d*a^3*A*tan(d*x+c)^(7/2
)-2/d*a^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-2/d*a^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-1
/d*a^3*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-2/9*I/d*a
^3*B*tan(d*x+c)^(9/2)-2*I/d*a^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*I/d*a^3*B*2^(1/2)*arctan(1+2^(1
/2)*tan(d*x+c)^(1/2))-1/d*a^3*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)
+tan(d*x+c)))-2/d*a^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-2/d*a^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x
+c)^(1/2))

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maxima [A]  time = 0.75, size = 232, normalized size = 1.17 \[ -\frac {70 i \, B a^{3} \tan \left (d x + c\right )^{\frac {9}{2}} - 2 \, {\left (-45 i \, A - 135 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {7}{2}} + 126 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {5}{2}} - 2 \, {\left (420 i \, A + 420 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {3}{2}} - 2520 \, {\left (A - i \, B\right )} a^{3} \sqrt {\tan \left (d x + c\right )} - 315 \, {\left (\sqrt {2} {\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(70*I*B*a^3*tan(d*x + c)^(9/2) - 2*(-45*I*A - 135*B)*a^3*tan(d*x + c)^(7/2) + 126*(3*A - 4*I*B)*a^3*tan
(d*x + c)^(5/2) - 2*(420*I*A + 420*B)*a^3*tan(d*x + c)^(3/2) - 2520*(A - I*B)*a^3*sqrt(tan(d*x + c)) - 315*(sq
rt(2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + sqrt(2)*(-(2*I + 2)*
A + (2*I - 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*((I - 1)*A + (I + 1)*B)*log(s
qrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)
) + tan(d*x + c) + 1))*a^3)/d

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mupad [B]  time = 10.95, size = 327, normalized size = 1.65 \[ \frac {8\,A\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,8{}\mathrm {i}}{3\,d}-\frac {6\,A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\frac {A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,2{}\mathrm {i}}{7\,d}-\frac {B\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,8{}\mathrm {i}}{d}+\frac {8\,B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,8{}\mathrm {i}}{5\,d}-\frac {6\,B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{7\,d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{9/2}\,2{}\mathrm {i}}{9\,d}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (-A\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,A\,a^3\,\ln \left (-A\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (-8\,B\,a^3\,d+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,B\,a^3\,\ln \left (-8\,B\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*A*a^3*tan(c + d*x)^(1/2))/d + (A*a^3*tan(c + d*x)^(3/2)*8i)/(3*d) - (6*A*a^3*tan(c + d*x)^(5/2))/(5*d) - (A
*a^3*tan(c + d*x)^(7/2)*2i)/(7*d) - (B*a^3*tan(c + d*x)^(1/2)*8i)/d + (8*B*a^3*tan(c + d*x)^(3/2))/(3*d) + (B*
a^3*tan(c + d*x)^(5/2)*8i)/(5*d) - (6*B*a^3*tan(c + d*x)^(7/2))/(7*d) - (B*a^3*tan(c + d*x)^(9/2)*2i)/(9*d) +
(2^(1/2)*A*a^3*log(- A*a^3*d*8i - 2^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - ((-16i)^(1/2)*A*a
^3*log(2*(-16i)^(1/2)*A*a^3*d*tan(c + d*x)^(1/2) - A*a^3*d*8i))/d + (2^(1/2)*B*a^3*log(- 8*B*a^3*d - 2^(1/2)*B
*a^3*d*tan(c + d*x)^(1/2)*(4 + 4i))*(2 + 2i))/d - (16i^(1/2)*B*a^3*log(2*16i^(1/2)*B*a^3*d*tan(c + d*x)^(1/2)
- 8*B*a^3*d))/d

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

[Out]

Timed out

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