3.340 \(\int \frac{\sec ^9(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=147 \[ \frac{8 i a^2 \sec ^9(c+d x)}{65 d (a+i a \tan (c+d x))^{5/2}}+\frac{64 i a^3 \sec ^9(c+d x)}{715 d (a+i a \tan (c+d x))^{7/2}}+\frac{256 i a^4 \sec ^9(c+d x)}{6435 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(((256*I)/6435)*a^4*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((64*I)/715)*a^3*Sec[c + d*x]^9)/(d*(a
+ I*a*Tan[c + d*x])^(7/2)) + (((8*I)/65)*a^2*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((2*I)/15)*a
*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(3/2))

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Rubi [A]  time = 0.258464, antiderivative size = 147, normalized size of antiderivative = 1., 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} \[ \frac{8 i a^2 \sec ^9(c+d x)}{65 d (a+i a \tan (c+d x))^{5/2}}+\frac{64 i a^3 \sec ^9(c+d x)}{715 d (a+i a \tan (c+d x))^{7/2}}+\frac{256 i a^4 \sec ^9(c+d x)}{6435 d (a+i a \tan (c+d x))^{9/2}}+\frac{2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((256*I)/6435)*a^4*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((64*I)/715)*a^3*Sec[c + d*x]^9)/(d*(a
+ I*a*Tan[c + d*x])^(7/2)) + (((8*I)/65)*a^2*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((2*I)/15)*a
*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(3/2))

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 \frac{\sec ^9(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{5} (4 a) \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\\ &=\frac{8 i a^2 \sec ^9(c+d x)}{65 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{65} \left (32 a^2\right ) \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{64 i a^3 \sec ^9(c+d x)}{715 d (a+i a \tan (c+d x))^{7/2}}+\frac{8 i a^2 \sec ^9(c+d x)}{65 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{715} \left (128 a^3\right ) \int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\\ &=\frac{256 i a^4 \sec ^9(c+d x)}{6435 d (a+i a \tan (c+d x))^{9/2}}+\frac{64 i a^3 \sec ^9(c+d x)}{715 d (a+i a \tan (c+d x))^{7/2}}+\frac{8 i a^2 \sec ^9(c+d x)}{65 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.501341, size = 95, normalized size = 0.65 \[ \frac{2 \sec ^8(c+d x) (3 i (90 \sin (c+d x)+233 \sin (3 (c+d x)))+510 \cos (c+d x)+731 \cos (3 (c+d x))) (\sin (4 (c+d x))+i \cos (4 (c+d x)))}{6435 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]^8*(510*Cos[c + d*x] + 731*Cos[3*(c + d*x)] + (3*I)*(90*Sin[c + d*x] + 233*Sin[3*(c + d*x)]))*(
I*Cos[4*(c + d*x)] + Sin[4*(c + d*x)]))/(6435*d*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 1.273, size = 154, normalized size = 1.1 \begin{align*}{\frac{4096\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+4096\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}-512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+1536\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -160\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1120\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -84\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+924\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -858\,i}{6435\,ad \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)^9/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/6435/d/a*(2048*I*cos(d*x+c)^8+2048*sin(d*x+c)*cos(d*x+c)^7-256*I*cos(d*x+c)^6+768*cos(d*x+c)^5*sin(d*x+c)-80
*I*cos(d*x+c)^4+560*cos(d*x+c)^3*sin(d*x+c)-42*I*cos(d*x+c)^2+462*cos(d*x+c)*sin(d*x+c)-429*I)*(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 [B]  time = 2.05545, size = 821, normalized size = 5.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/6435*(-1241*I*sqrt(a) - 5194*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 6090*I*sqrt(a)*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 + 2490*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14430*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 33618*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 13442*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) +
1)^6 - 18590*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 18590*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9
- 13442*I*sqrt(a)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 33618*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11
+ 14430*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 2490*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 -
6090*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 5194*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + 1
241*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16)*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*sqrt(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/((a - 8*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 - 56*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 56*a*sin(d*x + c
)^10/(cos(d*x + c) + 1)^10 + 28*a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a*sin(d*x + c)^14/(cos(d*x + c) +
1)^14 + a*sin(d*x + c)^16/(cos(d*x + c) + 1)^16)*d*sqrt(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/
(cos(d*x + c) + 1)^2 - 1))

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Fricas [A]  time = 2.10861, size = 529, normalized size = 3.6 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (183040 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 99840 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 30720 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4096 i\right )} e^{\left (i \, d x + i \, c\right )}}{6435 \,{\left (a d e^{\left (15 i \, d x + 15 i \, c\right )} + 7 \, a d e^{\left (13 i \, d x + 13 i \, c\right )} + 21 \, a d e^{\left (11 i \, d x + 11 i \, c\right )} + 35 \, a d e^{\left (9 i \, d x + 9 i \, c\right )} + 35 \, a d e^{\left (7 i \, d x + 7 i \, c\right )} + 21 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 7 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a 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)^9/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/6435*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(183040*I*e^(6*I*d*x + 6*I*c) + 99840*I*e^(4*I*d*x + 4*I*c) +
30720*I*e^(2*I*d*x + 2*I*c) + 4096*I)*e^(I*d*x + I*c)/(a*d*e^(15*I*d*x + 15*I*c) + 7*a*d*e^(13*I*d*x + 13*I*c
) + 21*a*d*e^(11*I*d*x + 11*I*c) + 35*a*d*e^(9*I*d*x + 9*I*c) + 35*a*d*e^(7*I*d*x + 7*I*c) + 21*a*d*e^(5*I*d*x
+ 5*I*c) + 7*a*d*e^(3*I*d*x + 3*I*c) + a*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)**9/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{9}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^9/sqrt(I*a*tan(d*x + c) + a), x)