3.326 \(\int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=139 \[ \frac{256 i a^4 \sec (c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{64 i a^3 \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac{2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \]

[Out]

(((256*I)/35)*a^4*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((64*I)/35)*a^3*Sec[c + d*x]*Sqrt[a + I*a*Ta
n[c + d*x]])/d + (((24*I)/35)*a^2*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*I)/7)*a*Sec[c + d*x]*(a
+ I*a*Tan[c + d*x])^(5/2))/d

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Rubi [A]  time = 0.139905, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3494, 3493} \[ \frac{256 i a^4 \sec (c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{64 i a^3 \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac{2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((256*I)/35)*a^4*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((64*I)/35)*a^3*Sec[c + d*x]*Sqrt[a + I*a*Ta
n[c + d*x]])/d + (((24*I)/35)*a^2*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*I)/7)*a*Sec[c + d*x]*(a
+ I*a*Tan[c + d*x])^(5/2))/d

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]

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]

Rubi steps

\begin{align*} \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=\frac{2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{7} (12 a) \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac{24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac{2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{35} \left (96 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac{64 i a^3 \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac{2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{35} \left (128 a^3\right ) \int \sec (c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{256 i a^4 \sec (c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{64 i a^3 \sec (c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac{2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.641958, size = 109, normalized size = 0.78 \[ \frac{2 a^3 \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (\sin (c-2 d x)+i \cos (c-2 d x)) (102 \cos (2 (c+d x))+14 i \tan (c+d x)+19 i \sin (3 (c+d x)) \sec (c+d x)+75)}{35 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*a^3*Sec[c + d*x]^2*(I*Cos[c - 2*d*x] + Sin[c - 2*d*x])*(75 + 102*Cos[2*(c + d*x)] + (19*I)*Sec[c + d*x]*Sin
[3*(c + d*x)] + (14*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [A]  time = 0.253, size = 100, normalized size = 0.7 \begin{align*}{\frac{2\,{a}^{3} \left ( 128\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+128\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +54\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-22\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -5\,i \right ) }{35\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

2/35/d*a^3*(128*I*cos(d*x+c)^4+128*cos(d*x+c)^3*sin(d*x+c)+54*I*cos(d*x+c)^2-22*cos(d*x+c)*sin(d*x+c)-5*I)*(a*
(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*sec(d*x + c), x)

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Fricas [A]  time = 2.04188, size = 363, normalized size = 2.61 \begin{align*} \frac{\sqrt{2}{\left (560 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 1120 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 896 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 256 i \, a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{35 \,{\left (d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/35*sqrt(2)*(560*I*a^3*e^(6*I*d*x + 6*I*c) + 1120*I*a^3*e^(4*I*d*x + 4*I*c) + 896*I*a^3*e^(2*I*d*x + 2*I*c) +
256*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^(7*I*d*x + 7*I*c) + 3*d*e^(5*I*d*x + 5*I*c)
+ 3*d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*sec(d*x + c), x)