3.287 \(\int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=110 \[ \frac {64 i a^3 \sec ^5(c+d x)}{315 d (a+i a \tan (c+d x))^{5/2}}+\frac {16 i a^2 \sec ^5(c+d x)}{63 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2/9*I*a*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(1/2)+64/315*I*a^3*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(5/2)+16/63*I*a
^2*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.18, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac {16 i a^2 \sec ^5(c+d x)}{63 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^5(c+d x)}{315 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((64*I)/315)*a^3*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((16*I)/63)*a^2*Sec[c + d*x]^5)/(d*(a +
I*a*Tan[c + d*x])^(3/2)) + (((2*I)/9)*a*Sec[c + d*x]^5)/(d*Sqrt[a + I*a*Tan[c + d*x]])

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 ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{9} (8 a) \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {16 i a^2 \sec ^5(c+d x)}{63 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{63} \left (32 a^2\right ) \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\\ &=\frac {64 i a^3 \sec ^5(c+d x)}{315 d (a+i a \tan (c+d x))^{5/2}}+\frac {16 i a^2 \sec ^5(c+d x)}{63 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 77, normalized size = 0.70 \[ \frac {2 \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (55 i \sin (2 (c+d x))+71 \cos (2 (c+d x))+36) (\sin (3 (c+d x))+i \cos (3 (c+d x)))}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]^4*(36 + 71*Cos[2*(c + d*x)] + (55*I)*Sin[2*(c + d*x)])*(I*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])
*Sqrt[a + I*a*Tan[c + d*x]])/(315*d)

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fricas [A]  time = 1.46, size = 97, normalized size = 0.88 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2016 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1152 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 256 i\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(2016*I*e^(4*I*d*x + 4*I*c) + 1152*I*e^(2*I*d*x + 2*I*c) + 256
*I)/(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 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 1.28, size = 114, normalized size = 1.04 \[ \frac {2 \left (128 i \left (\cos ^{5}\left (d x +c \right )\right )+128 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-16 i \left (\cos ^{3}\left (d x +c \right )\right )+48 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 i \cos \left (d x +c \right )+35 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{315 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/315/d*(128*I*cos(d*x+c)^5+128*sin(d*x+c)*cos(d*x+c)^4-16*I*cos(d*x+c)^3+48*cos(d*x+c)^2*sin(d*x+c)-5*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

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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)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 6.08, size = 102, normalized size = 0.93 \[ \frac {32\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,36{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,63{}\mathrm {i}+8{}\mathrm {i}\right )}{315\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(32*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)*36i + exp(c*4i + d*x*4i)*63i + 8i))/(315*d*(exp(c*2i + d*x*2i) + 1)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

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

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