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

Optimal. Leaf size=104 \[ \frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \]

[Out]

64/15*I*a^3*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)+16/15*I*a^2*sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d+2/5*I*a*se
c(d*x+c)*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]  time = 0.10, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3494, 3493} \[ \frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((64*I)/15)*a^3*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*a^2*Sec[c + d*x]*Sqrt[a + I*a*Tan
[c + d*x]])/d + (((2*I)/5)*a*Sec[c + d*x]*(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 (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{5} (8 a) \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{15} \left (32 a^2\right ) \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 93, normalized size = 0.89 \[ \frac {2 a^2 \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (\sin (c-d x)+i \cos (c-d x)) (7 i \sin (2 (c+d x))+23 \cos (2 (c+d x))+20)}{15 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sec[c + d*x]^2*(I*Cos[c - d*x] + Sin[c - d*x])*(20 + 23*Cos[2*(c + d*x)] + (7*I)*Sin[2*(c + d*x)])*Sqrt
[a + I*a*Tan[c + d*x]])/(15*d*(Cos[d*x] + I*Sin[d*x])^2)

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fricas [A]  time = 0.62, size = 83, normalized size = 0.80 \[ \frac {\sqrt {2} {\left (120 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 160 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*(120*I*a^2*e^(4*I*d*x + 4*I*c) + 160*I*a^2*e^(2*I*d*x + 2*I*c) + 64*I*a^2)*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1))/(d*e^(4*I*d*x + 4*I*c) + 2*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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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maple [A]  time = 0.90, size = 90, normalized size = 0.87 \[ \frac {2 \left (32 i \left (\cos ^{3}\left (d x +c \right )\right )+32 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+11 i \cos \left (d x +c \right )-3 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{15 d \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(32*I*cos(d*x+c)^3+32*cos(d*x+c)^2*sin(d*x+c)+11*I*cos(d*x+c)-3*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c)
)/cos(d*x+c))^(1/2)/cos(d*x+c)^2*a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.08, size = 105, normalized size = 1.01 \[ \frac {8\,a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,20{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,15{}\mathrm {i}+8{}\mathrm {i}\right )}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

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),x)

[Out]

(8*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)*(exp(c*2i
+ d*x*2i)*20i + exp(c*4i + d*x*4i)*15i + 8i))/(15*d*(exp(c*2i + d*x*2i) + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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