3.4.33 \(\int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [333]

Optimal. Leaf size=117 \[ -\frac {16 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{9/2}}{3 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{11/2}}{11 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^7 d} \]

[Out]

-16/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^4/d+8/3*I*(a+I*a*tan(d*x+c))^(9/2)/a^5/d-12/11*I*(a+I*a*tan(d*x+c))^(11/2)/
a^6/d+2/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^7/d

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Rubi [A]
time = 0.06, antiderivative size = 117, 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 = {3568, 45} \begin {gather*} \frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^7 d}-\frac {12 i (a+i a \tan (c+d x))^{11/2}}{11 a^6 d}+\frac {8 i (a+i a \tan (c+d x))^{9/2}}{3 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-16*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^4*d) + (((8*I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^5*d) - (((12*
I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^6*d) + (((2*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^7*d)

Rule 45

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])

Rule 3568

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]

Rubi steps

\begin {align*} \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {i \text {Subst}\left (\int (a-x)^3 (a+x)^{5/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {i \text {Subst}\left (\int \left (8 a^3 (a+x)^{5/2}-12 a^2 (a+x)^{7/2}+6 a (a+x)^{9/2}-(a+x)^{11/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {16 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{9/2}}{3 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{11/2}}{11 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^7 d}\\ \end {align*}

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Mathematica [A]
time = 0.68, size = 95, normalized size = 0.81 \begin {gather*} \frac {2 \sec ^7(c+d x) (390 \cos (c+d x)+445 \cos (3 (c+d x))-7 i (26 \sin (c+d x)+59 \sin (3 (c+d x)))) (-i \cos (4 (c+d x))+\sin (4 (c+d x)))}{3003 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]^7*(390*Cos[c + d*x] + 445*Cos[3*(c + d*x)] - (7*I)*(26*Sin[c + d*x] + 59*Sin[3*(c + d*x)]))*((
-I)*Cos[4*(c + d*x)] + Sin[4*(c + d*x)]))/(3003*d*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.96, size = 127, normalized size = 1.09

method	result	size
default	\(-\frac {2 \left (512 i \left (\cos ^{6}\left (d x +c \right )\right )-512 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+64 i \left (\cos ^{4}\left (d x +c \right )\right )-320 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+28 i \left (\cos ^{2}\left (d x +c \right )\right )-252 \sin \left (d x +c \right ) \cos \left (d x +c \right )+231 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3003 d \cos \left (d x +c \right )^{6} a}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3003/d*(512*I*cos(d*x+c)^6-512*sin(d*x+c)*cos(d*x+c)^5+64*I*cos(d*x+c)^4-320*sin(d*x+c)*cos(d*x+c)^3+28*I*c
os(d*x+c)^2-252*sin(d*x+c)*cos(d*x+c)+231*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^6/a

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (85) = 170\).
time = 0.28, size = 297, normalized size = 2.54 \begin {gather*} -\frac {2 i \, {\left (15015 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} - \frac {3003 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{a^{2}} + \frac {143 \, {\left (35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 180 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 378 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}\right )}}{a^{4}} - \frac {5 \, {\left (231 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 1638 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{a^{6}}\right )}}{15015 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/15015*I*(15015*sqrt(I*a*tan(d*x + c) + a) - 3003*(3*(I*a*tan(d*x + c) + a)^(5/2) - 10*(I*a*tan(d*x + c) + a
)^(3/2)*a + 15*sqrt(I*a*tan(d*x + c) + a)*a^2)/a^2 + 143*(35*(I*a*tan(d*x + c) + a)^(9/2) - 180*(I*a*tan(d*x +
c) + a)^(7/2)*a + 378*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 420*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 315*sqrt(I*a*
tan(d*x + c) + a)*a^4)/a^4 - 5*(231*(I*a*tan(d*x + c) + a)^(13/2) - 1638*(I*a*tan(d*x + c) + a)^(11/2)*a + 500
5*(I*a*tan(d*x + c) + a)^(9/2)*a^2 - 8580*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 9009*(I*a*tan(d*x + c) + a)^(5/2)
*a^4 - 6006*(I*a*tan(d*x + c) + a)^(3/2)*a^5 + 3003*sqrt(I*a*tan(d*x + c) + a)*a^6)/a^6)/(a*d)

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Fricas [A]
time = 0.39, size = 150, normalized size = 1.28 \begin {gather*} -\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (13 i \, d x + 13 i \, c\right )} + 104 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 286 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 429 i \, e^{\left (7 i \, d x + 7 i \, c\right )}\right )}}{3003 \, {\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-128/3003*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(16*I*e^(13*I*d*x + 13*I*c) + 104*I*e^(11*I*d*x + 11*I*c)
+ 286*I*e^(9*I*d*x + 9*I*c) + 429*I*e^(7*I*d*x + 7*I*c))/(a*d*e^(12*I*d*x + 12*I*c) + 6*a*d*e^(10*I*d*x + 10*I
*c) + 15*a*d*e^(8*I*d*x + 8*I*c) + 20*a*d*e^(6*I*d*x + 6*I*c) + 15*a*d*e^(4*I*d*x + 4*I*c) + 6*a*d*e^(2*I*d*x
+ 2*I*c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{8}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**8/sqrt(I*a*(tan(c + d*x) - I)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^8/sqrt(I*a*tan(d*x + c) + a), x)

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Mupad [B]
time = 8.94, size = 434, normalized size = 3.71 \begin {gather*} -\frac {\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}}\,2048{}\mathrm {i}}{3003\,a\,d}-\frac {\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}}\,1024{}\mathrm {i}}{3003\,a\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\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}}{1001\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\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}}\,640{}\mathrm {i}}{3003\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\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}}\,6784{}\mathrm {i}}{429\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\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}}\,3456{}\mathrm {i}}{143\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {\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}}\,128{}\mathrm {i}}{13\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*6784i)/(429*a*d*(exp(c*2i + d*x*2i)
+ 1)^4) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*1024i)/(3003*a*d*(exp(c*2i
+ d*x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*256i)/(1001*a*d*(
exp(c*2i + d*x*2i) + 1)^2) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*640i)/(
3003*a*d*(exp(c*2i + d*x*2i) + 1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/
2)*2048i)/(3003*a*d) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*3456i)/(143*a
*d*(exp(c*2i + d*x*2i) + 1)^5) + ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128
i)/(13*a*d*(exp(c*2i + d*x*2i) + 1)^6)

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