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

Optimal. Leaf size=183 \[ -\frac {4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {4 a^2 (B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {4 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

[Out]

-4*(-1)^(1/4)*a^2*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-4*a^2*(I*A+B)*tan(d*x+c)^(1/2)/d+4/3*a^2*(A-I*
B)*tan(d*x+c)^(3/2)/d+4/5*a^2*(I*A+B)*tan(d*x+c)^(5/2)/d-2/63*a^2*(9*A-11*I*B)*tan(d*x+c)^(7/2)/d+2/9*I*B*tan(
d*x+c)^(7/2)*(a^2+I*a^2*tan(d*x+c))/d

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Rubi [A]  time = 0.36, antiderivative size = 183, 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} \[ -\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {4 a^2 (B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {4 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (4*a^2*(I*A + B)*Sqrt[Tan[c + d*x]])/d
+ (4*a^2*(A - I*B)*Tan[c + d*x]^(3/2))/(3*d) + (4*a^2*(I*A + B)*Tan[c + d*x]^(5/2))/(5*d) - (2*a^2*(9*A - (11
*I)*B)*Tan[c + d*x]^(7/2))/(63*d) + (((2*I)/9)*B*Tan[c + d*x]^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))/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 {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \left (\frac {1}{2} a (9 A-7 i B)+\frac {1}{2} a (9 i A+11 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \tan ^{\frac {5}{2}}(c+d x) \left (9 a^2 (A-i B)+9 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \tan ^{\frac {3}{2}}(c+d x) \left (-9 a^2 (i A+B)+9 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \sqrt {\tan (c+d x)} \left (-9 a^2 (A-i B)-9 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=-\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \frac {9 a^2 (i A+B)-9 a^2 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {\left (36 a^4 (i A+B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{9 a^2 (i A+B)+9 a^2 (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {4 \sqrt [4]{-1} a^2 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}\\ \end {align*}

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Mathematica [A]  time = 6.94, size = 315, normalized size = 1.72 \[ \frac {\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (\frac {4 e^{-2 i c} (B+i A) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {i (\cos (2 c)-i \sin (2 c)) \sqrt {\tan (c+d x)} \sec ^4(c+d x) (30 (8 B+11 i A) \sin (2 (c+d x))+15 (20 B+17 i A) \sin (4 (c+d x))+140 (18 A-17 i B) \cos (2 (c+d x))+(756 A-791 i B) \cos (4 (c+d x))+21 (84 A-89 i B))}{1260}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^3*((4*(I*A + 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)))]])/(E^((2*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]) - (I/1260)*Sec[c + d*x]^4*(Cos[2*c] - I*Sin[2*c])*(21*(84*A - (89*I)*B) + 140*(18*A - (17*I)
*B)*Cos[2*(c + d*x)] + (756*A - (791*I)*B)*Cos[4*(c + d*x)] + 30*((11*I)*A + 8*B)*Sin[2*(c + d*x)] + 15*((17*I
)*A + 20*B)*Sin[4*(c + d*x)])*Sqrt[Tan[c + d*x]])*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*(Cos[d*x]
+ I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.77, size = 547, normalized size = 2.99 \[ \frac {315 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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 (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 315 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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 (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) + {\left ({\left (-8088 i \, A - 8728 \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-22800 i \, A - 20960 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-28224 i \, A - 29904 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-17520 i \, A - 17120 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-4008 i \, A - 3928 \, B\right )} a^{2}\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)^(5/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/1260*(315*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + sqrt((16*I*A^2
+ 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + 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)/((2*I*A + 2*B)*a^2)) - 315*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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((4*(A
- I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + 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)/((2*I*A + 2*B)*a^2)) + ((-8088*
I*A - 8728*B)*a^2*e^(8*I*d*x + 8*I*c) + (-22800*I*A - 20960*B)*a^2*e^(6*I*d*x + 6*I*c) + (-28224*I*A - 29904*B
)*a^2*e^(4*I*d*x + 4*I*c) + (-17520*I*A - 17120*B)*a^2*e^(2*I*d*x + 2*I*c) + (-4008*I*A - 3928*B)*a^2)*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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \tan \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.10, size = 607, normalized size = 3.32 \[ -\frac {2 a^{2} B \left (\tan ^{\frac {9}{2}}\left (d x +c \right )\right )}{9 d}+\frac {i a^{2} 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 )}{2 d}-\frac {2 a^{2} A \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7 d}-\frac {4 i a^{2} B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {4 a^{2} B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {i a^{2} 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 )}{2 d}+\frac {4 a^{2} A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {4 i a^{2} B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7 d}-\frac {4 a^{2} B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {i a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {4 i a^{2} A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {i a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {a^{2} 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 )}{2 d}+\frac {a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {4 i a^{2} A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {i a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {i a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} 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 )}{2 d}-\frac {a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} A \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)^(5/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

-2/9/d*a^2*B*tan(d*x+c)^(9/2)+1/2*I/d*a^2*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)))-2/7/d*a^2*A*tan(d*x+c)^(7/2)-4/3*I/d*a^2*B*tan(d*x+c)^(3/2)+4/5/d*a^2*B*tan(d*x+c)^(
5/2)+1/2*I/d*a^2*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)))
+4/3/d*a^2*A*tan(d*x+c)^(3/2)+4/7*I/d*a^2*B*tan(d*x+c)^(7/2)-4/d*a^2*B*tan(d*x+c)^(1/2)+I/d*a^2*A*2^(1/2)*arct
an(1+2^(1/2)*tan(d*x+c)^(1/2))+4/5*I/d*a^2*A*tan(d*x+c)^(5/2)+I/d*a^2*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(
1/2))+1/d*a^2*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+1/2/d*a^2*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)))+1/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-4*I
/d*a^2*A*tan(d*x+c)^(1/2)+I/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+I/d*a^2*A*2^(1/2)*arctan(-1+2^(
1/2)*tan(d*x+c)^(1/2))-1/2/d*a^2*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)))-1/d*a^2*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-1/d*a^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(
d*x+c)^(1/2))

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maxima [A]  time = 0.74, size = 232, normalized size = 1.27 \[ -\frac {140 \, B a^{2} \tan \left (d x + c\right )^{\frac {9}{2}} + 180 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {7}{2}} - 4 \, {\left (126 i \, A + 126 \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {5}{2}} - 840 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 4 \, {\left (-630 i \, A - 630 \, B\right )} a^{2} \sqrt {\tan \left (d x + c\right )} - 315 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2}}{630 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/630*(140*B*a^2*tan(d*x + c)^(9/2) + 180*(A - 2*I*B)*a^2*tan(d*x + c)^(7/2) - 4*(126*I*A + 126*B)*a^2*tan(d*
x + c)^(5/2) - 840*(A - I*B)*a^2*tan(d*x + c)^(3/2) - 4*(-630*I*A - 630*B)*a^2*sqrt(tan(d*x + c)) - 315*(2*sqr
t(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I
+ 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(sqrt(2)*s
qrt(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^2)/d

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mupad [B]  time = 11.03, size = 326, normalized size = 1.78 \[ \frac {4\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}-\frac {A\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,4{}\mathrm {i}}{d}+\frac {A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,4{}\mathrm {i}}{5\,d}-\frac {2\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{7\,d}-\frac {4\,B\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}-\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}+\frac {4\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}+\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,4{}\mathrm {i}}{7\,d}-\frac {2\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{9\,d}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (-4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (-4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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