3.313 \(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=147 \[ \frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

[Out]

64/105*I*a^3*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)+8/21*I*a^2*sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d+256/31
5*I*a^4*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)+2/9*I*a*sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]  time = 0.24, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((256*I)/315)*a^4*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*Sqr
t[a + I*a*Tan[c + d*x]]) + (((8*I)/21)*a^2*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (((2*I)/9)*a*Sec[c +
d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d

Rule 3493

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*
(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3494

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {1}{3} (4 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {1}{21} \left (32 a^2\right ) \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {1}{105} \left (128 a^3\right ) \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.84, size = 103, normalized size = 0.70 \[ \frac {2 a^2 (\sin (2 c)+i \cos (2 c)) \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (242 \cos (2 (c+d x))+54 i \tan (c+d x)+89 i \sin (3 (c+d x)) \sec (c+d x)+77)}{315 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sec[c + d*x]^3*(I*Cos[2*c] + Sin[2*c])*(77 + 242*Cos[2*(c + d*x)] + (89*I)*Sec[c + d*x]*Sin[3*(c + d*x)
] + (54*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*(Cos[d*x] + I*Sin[d*x])^2)

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fricas [A]  time = 0.60, size = 121, normalized size = 0.82 \[ \frac {\sqrt {2} {\left (3360 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 4032 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2304 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 512 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*sqrt(2)*(3360*I*a^2*e^(6*I*d*x + 6*I*c) + 4032*I*a^2*e^(4*I*d*x + 4*I*c) + 2304*I*a^2*e^(2*I*d*x + 2*I*c
) + 512*I*a^2)*sqrt(a/(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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.19, size = 117, normalized size = 0.80 \[ \frac {2 \left (256 i \left (\cos ^{5}\left (d x +c \right )\right )+256 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-32 i \left (\cos ^{3}\left (d x +c \right )\right )+96 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+95 i \cos \left (d x +c \right )-35 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{315 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315/d*(256*I*cos(d*x+c)^5+256*sin(d*x+c)*cos(d*x+c)^4-32*I*cos(d*x+c)^3+96*cos(d*x+c)^2*sin(d*x+c)+95*I*cos(
d*x+c)-35*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4*a^2

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maxima [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(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 7.77, size = 301, normalized size = 2.05 \[ \frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{3\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(3*d*(
exp(c*2i + d*x*2i) + 1)) - (a^2*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*
2i) + 1))^(1/2)*96i)/(5*d*(exp(c*2i + d*x*2i) + 1)^2) + (a^2*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*
1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*96i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3) - (a^2*exp(- c*1i - d*x*1i)
*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(9*d*(exp(c*2i + d*x*2i) + 1)^4
)

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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(sec(d*x+c)**3*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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