\(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx\) [177]

Optimal result

Integrand size = 38, antiderivative size = 323 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\frac {(4+4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (2155 i A+2167 B) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \]

[Out]

(4+4*I)*a^(5/2)*(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-8/3465*a^2*(2155*I*
A+2167*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/99*a^2*(14*I*A+11*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*
x+c)^(9/2)+2/693*a^2*(212*A-209*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)+4/1155*a^2*(250*I*A+253*B)*(a
+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-8/3465*a^2*(655*A-649*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/
2)-2/11*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(11/2)

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3674, 3679, 12, 3625, 211} \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\frac {(4+4 i) a^{5/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {8 a^2 (2167 B+2155 i A) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(13/2),x]

[Out]

((4 + 4*I)*a^(5/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a^2*((14*I)*A + 11*B)*Sqrt[a + I*a*Tan[c + d*x]])/(99*d*Tan[c + d*x]^(9/2)) + (2*a^2*(212*A - (209*I)*B)*Sqrt[
a + I*a*Tan[c + d*x]])/(693*d*Tan[c + d*x]^(7/2)) + (4*a^2*((250*I)*A + 253*B)*Sqrt[a + I*a*Tan[c + d*x]])/(11
55*d*Tan[c + d*x]^(5/2)) - (8*a^2*(655*A - (649*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3465*d*Tan[c + d*x]^(3/2))
- (8*a^2*((2155*I)*A + 2167*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3465*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c
+ d*x])^(3/2))/(11*d*Tan[c + d*x]^(11/2))

Rule 12

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3625

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 3674

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

Rule 3679

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*} \text {integral}& = -\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {2}{11} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (14 i A+11 B)-\frac {1}{2} a (8 A-11 i B) \tan (c+d x)\right )}{\tan ^{\frac {11}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {4}{99} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (212 A-209 i B)-\frac {1}{4} a^2 (184 i A+187 B) \tan (c+d x)\right )}{\tan ^{\frac {9}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^3 (250 i A+253 B)+\frac {3}{4} a^3 (212 A-209 i B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx}{693 a} \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^4 (655 A-649 i B)+\frac {3}{2} a^4 (250 i A+253 B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{3465 a^2} \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {32 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{8} a^5 (2155 i A+2167 B)-\frac {3}{4} a^5 (655 A-649 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{10395 a^3} \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (2155 i A+2167 B) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {64 \int -\frac {10395 a^6 (A-i B) \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{10395 a^4} \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (2155 i A+2167 B) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}-\left (4 a^2 (A-i B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (2155 i A+2167 B) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {\left (8 a^4 (i A+B)\right ) \text {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 {(4+4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (2155 i A+2167 B) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.71 (sec) , antiderivative size = 613, normalized size of antiderivative = 1.90 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=-\frac {2 A (a+i a \tan (c+d x))^{5/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {2 \left (-\frac {a (5 i A+11 B) (a+i a \tan (c+d x))^{5/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (\frac {a^2 (79 A-55 i B) (a+i a \tan (c+d x))^{5/2}}{14 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (\frac {a^3 (535 i A+583 B) (a+i a \tan (c+d x))^{5/2}}{20 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {693}{8} a^3 (A-i B) \left (\frac {4 i \sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {4 i a^{5/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 (-1)^{3/4} a^2 \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)}}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 i a^2 \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}}\right )\right )}{7 a}\right )}{9 a}\right )}{11 a} \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(13/2),x]

[Out]

(-2*A*(a + I*a*Tan[c + d*x])^(5/2))/(11*d*Tan[c + d*x]^(11/2)) + (2*(-1/9*(a*((5*I)*A + 11*B)*(a + I*a*Tan[c +
d*x])^(5/2))/(d*Tan[c + d*x]^(9/2)) + (2*((a^2*(79*A - (55*I)*B)*(a + I*a*Tan[c + d*x])^(5/2))/(14*d*Tan[c +
d*x]^(7/2)) + (2*((a^3*((535*I)*A + 583*B)*(a + I*a*Tan[c + d*x])^(5/2))/(20*d*Tan[c + d*x]^(5/2)) + (693*a^3*
(A - I*B)*(((4*I)*Sqrt[2]*a^2*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[I*a*Ta
n[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - ((4*I)*a^(5/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c
+ d*x]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (5*(-1)^(3/4)*a^2*ArcSinh[
(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[1 + I*Tan[c + d*x]]) - (2*a^2*Sqrt[a + I*a*
Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (((14*I)/3)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) -
(I*a^(3/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt
[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])))/8))/(7*a)))/(9*a)))/(11*a)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 973 vs. \(2 (266 ) = 532\).

Time = 0.15 (sec) , antiderivative size = 974, normalized size of antiderivative = 3.02

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(974\)
default	\(\text {Expression too large to display}\)	\(974\)
parts	\(\text {Expression too large to display}\)	\(1045\)

[In]

int((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x,method=_RETURNVERBOSE)

[Out]

-1/3465/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(11/2)*(17336*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^5*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-13860*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(
I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6-5192*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)+5240*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/
2)+2090*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+13860*A*ln(1/2*(2*I*a*
tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6+161
0*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-3465*(I*a)^(1/2)*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))*a*tan(d*x+c)^6
-3036*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+17240*I*A*(I*a)^(1/2)*(-I*
a)^(1/2)*tan(d*x+c)^5*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+6930*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6-2120*A*(I*a)^(1/2)*(-I*a)^(1/2)*ta
n(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+6930*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+
c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6+3465*I*(I*a)^(1/2)*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))*a*tan(d*x+c)^6-3000*I*A*(
I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+770*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+630*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 771 vs. \(2 (249) = 498\).

Time = 0.27 (sec) , antiderivative size = 771, normalized size of antiderivative = 2.39 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/3465*(3465*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*
c) + 15*d*e^(8*I*d*x + 8*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) - 6*d*e^(2*I*d*x + 2*I*c)
+ d)*log((I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*
d*x + 2*I*c) + (-I*A - B)*a^2)*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)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 3465*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e
^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*
e^(4*I*d*x + 4*I*c) - 6*d*e^(2*I*d*x + 2*I*c) + d)*log((-I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*d*e^
(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^2)*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)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) + 2*sqrt(2)
*(2*(3730*A - 3553*I*B)*a^2*e^(13*I*d*x + 13*I*c) - 9*(1805*A - 2013*I*B)*a^2*e^(11*I*d*x + 11*I*c) + 55*(397*
A - 337*I*B)*a^2*e^(9*I*d*x + 9*I*c) + 66*(95*A - 47*I*B)*a^2*e^(7*I*d*x + 7*I*c) - 1386*(15*A - 16*I*B)*a^2*e
^(5*I*d*x + 5*I*c) + 15015*(A - I*B)*a^2*e^(3*I*d*x + 3*I*c) - 3465*(A - I*B)*a^2*e^(I*d*x + I*c))*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)))/(d*e^(12*I*d*x + 12*I*c)
- 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) -
6*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(13/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Non regular value [0] was discarded and replaced randomly by 0=[-42]Warning, replacing -42 by -80, a substi
tution vari

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{13/2}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(13/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(13/2), x)