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

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

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 147, normalized size of antiderivative = 1.00, 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 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 \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*}

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Mathematica [A]  time = 0.67, 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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fricas [A]  time = 0.79, size = 153, normalized size = 1.04 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (183040 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 99840 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 30720 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4096 i\right )}}{6435 \, {\left (a d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

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)/(a*d*e^(14*I*d*x + 14*I*c) + 7*a*d*e^(12*I*d*x + 12*I*c) + 21*a*d*e^(10
*I*d*x + 10*I*c) + 35*a*d*e^(8*I*d*x + 8*I*c) + 35*a*d*e^(6*I*d*x + 6*I*c) + 21*a*d*e^(4*I*d*x + 4*I*c) + 7*a*
d*e^(2*I*d*x + 2*I*c) + a*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{9}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

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)

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maple [A]  time = 2.96, size = 154, normalized size = 1.05 \[ \frac {2 \left (2048 i \left (\cos ^{8}\left (d x +c \right )\right )+2048 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-256 i \left (\cos ^{6}\left (d x +c \right )\right )+768 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-80 i \left (\cos ^{4}\left (d x +c \right )\right )+560 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-42 i \left (\cos ^{2}\left (d x +c \right )\right )+462 \cos \left (d x +c \right ) \sin \left (d x +c \right )-429 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{6435 d \cos \left (d x +c \right )^{7} a} \]

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*(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/a

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maxima [B]  time = 0.90, size = 608, normalized size = 4.14 \[ -\frac {2 \, {\left (-1241 i \, \sqrt {a} - \frac {5194 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6090 i \, \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2490 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14430 i \, \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {33618 \, \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {13442 i \, \sqrt {a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {18590 \, \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {18590 \, \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {13442 i \, \sqrt {a} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {33618 \, \sqrt {a} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + \frac {14430 i \, \sqrt {a} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {2490 \, \sqrt {a} \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {6090 i \, \sqrt {a} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} - \frac {5194 \, \sqrt {a} \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}} + \frac {1241 i \, \sqrt {a} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}\right )} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1}}{6435 \, {\left (a - \frac {8 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {28 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {56 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {70 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {56 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {28 \, a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {8 \, a \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {a \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}\right )} d \sqrt {-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \]

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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mupad [B]  time = 8.98, size = 301, normalized size = 2.05 \[ \frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{9\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,768{}\mathrm {i}}{11\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,768{}\mathrm {i}}{13\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{15\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

(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)*256i)/(9*a*d*(e
xp(c*2i + d*x*2i) + 1)^4) - (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)*768i)/(11*a*d*(exp(c*2i + d*x*2i) + 1)^5) + (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)*768i)/(13*a*d*(exp(c*2i + d*x*2i) + 1)^6) - (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)*256i)/(15*a*d*(exp(c*2i + d*x*2i) + 1)^
7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{9}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

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]

Integral(sec(c + d*x)**9/sqrt(I*a*(tan(c + d*x) - I)), x)

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