3.307 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d}+\frac{8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac{8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d} \]

[Out]

(((-8*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^3*d) + (((8*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^4*d) - (((
2*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(a^5*d)

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Rubi [A]  time = 0.0771402, antiderivative size = 88, normalized size of antiderivative = 1., 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 = {3487, 43} \[ -\frac{2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d}+\frac{8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac{8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-8*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^3*d) + (((8*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^4*d) - (((
2*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(a^5*d)

Rule 3487

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^2 (a+x)^{9/2} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (4 a^2 (a+x)^{9/2}-4 a (a+x)^{11/2}+(a+x)^{13/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}+\frac{8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac{2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d}\\ \end{align*}

Mathematica [A]  time = 0.745479, size = 97, normalized size = 1.1 \[ \frac{2 a^2 \sec ^7(c+d x) \sqrt{a+i a \tan (c+d x)} (-187 i \sin (2 (c+d x))+203 \cos (2 (c+d x))+60) (\sin (5 c+7 d x)-i \cos (5 c+7 d x))}{2145 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*a^2*Sec[c + d*x]^7*(60 + 203*Cos[2*(c + d*x)] - (187*I)*Sin[2*(c + d*x)])*((-I)*Cos[5*c + 7*d*x] + Sin[5*c
+ 7*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(2145*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [B]  time = 0.581, size = 144, normalized size = 1.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( 512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}-512\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) +64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}-320\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+28\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-252\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -341\,i\cos \left ( dx+c \right ) +143\,\sin \left ( dx+c \right ) \right ) }{2145\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}}\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)^6*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-2/2145/d*a^2*(512*I*cos(d*x+c)^7-512*cos(d*x+c)^6*sin(d*x+c)+64*I*cos(d*x+c)^5-320*sin(d*x+c)*cos(d*x+c)^4+28
*I*cos(d*x+c)^3-252*cos(d*x+c)^2*sin(d*x+c)-341*I*cos(d*x+c)+143*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(
d*x+c))^(1/2)/cos(d*x+c)^7

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Maxima [A]  time = 1.09649, size = 78, normalized size = 0.89 \begin{align*} -\frac{2 i \,{\left (143 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{15}{2}} - 660 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} a + 780 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a^{2}\right )}}{2145 \, a^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/2145*I*(143*(I*a*tan(d*x + c) + a)^(15/2) - 660*(I*a*tan(d*x + c) + a)^(13/2)*a + 780*(I*a*tan(d*x + c) + a
)^(11/2)*a^2)/(a^5*d)

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Fricas [B]  time = 2.61204, size = 497, normalized size = 5.65 \begin{align*} \frac{\sqrt{2}{\left (-2048 i \, a^{2} e^{\left (14 i \, d x + 14 i \, c\right )} - 15360 i \, a^{2} e^{\left (12 i \, d x + 12 i \, c\right )} - 49920 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{2145 \,{\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/2145*sqrt(2)*(-2048*I*a^2*e^(14*I*d*x + 14*I*c) - 15360*I*a^2*e^(12*I*d*x + 12*I*c) - 49920*I*a^2*e^(10*I*d*
x + 10*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^(14*I*d*x + 14*I*c) + 7*d*e^(12*I*d*x + 12
*I*c) + 21*d*e^(10*I*d*x + 10*I*c) + 35*d*e^(8*I*d*x + 8*I*c) + 35*d*e^(6*I*d*x + 6*I*c) + 21*d*e^(4*I*d*x + 4
*I*c) + 7*d*e^(2*I*d*x + 2*I*c) + d)

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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)**6*(a+I*a*tan(d*x+c))**(5/2),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{5}{2}} \sec \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(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)^6, x)