3.3.79 \(\int \sec ^8(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [279]

Optimal. Leaf size=117 \[ -\frac {16 i (a+i a \tan (c+d x))^{9/2}}{9 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{11/2}}{11 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{13/2}}{13 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^7 d} \]

[Out]

-16/9*I*(a+I*a*tan(d*x+c))^(9/2)/a^4/d+24/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^5/d-12/13*I*(a+I*a*tan(d*x+c))^(13/
2)/a^6/d+2/15*I*(a+I*a*tan(d*x+c))^(15/2)/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))^{15/2}}{15 a^7 d}-\frac {12 i (a+i a \tan (c+d x))^{13/2}}{13 a^6 d}+\frac {24 i (a+i a \tan (c+d x))^{11/2}}{11 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{9/2}}{9 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)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(a^4*d) + (((24*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^5*d) - (((
12*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^6*d) + (((2*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(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 \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)^{7/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {i \text {Subst}\left (\int \left (8 a^3 (a+x)^{7/2}-12 a^2 (a+x)^{9/2}+6 a (a+x)^{11/2}-(a+x)^{13/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {16 i (a+i a \tan (c+d x))^{9/2}}{9 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{11/2}}{11 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{13/2}}{13 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^7 d}\\ \end {align*}

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Mathematica [A]
time = 0.95, size = 95, normalized size = 0.81 \begin {gather*} \frac {2 \sec ^7(c+d x) (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))) \sqrt {a+i a \tan (c+d x)}}{6435 d} \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*(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)])*Sqrt[a + I*a*Tan[c + d*x]])/(6435*d)

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Maple [A]
time = 4.78, size = 141, normalized size = 1.21

method	result	size
default	\(-\frac {2 \left (1024 i \left (\cos ^{7}\left (d x +c \right )\right )-1024 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )+128 i \left (\cos ^{5}\left (d x +c \right )\right )-640 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+56 i \left (\cos ^{3}\left (d x +c \right )\right )-504 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+33 i \cos \left (d x +c \right )-429 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{6435 d \cos \left (d x +c \right )^{7}}\)	\(141\)

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]

-2/6435/d*(1024*I*cos(d*x+c)^7-1024*sin(d*x+c)*cos(d*x+c)^6+128*I*cos(d*x+c)^5-640*sin(d*x+c)*cos(d*x+c)^4+56*
I*cos(d*x+c)^3-504*cos(d*x+c)^2*sin(d*x+c)+33*I*cos(d*x+c)-429*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 = 0.28, size = 76, normalized size = 0.65 \begin {gather*} \frac {2 i \, {\left (429 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 2970 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 7020 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} - 5720 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}\right )}}{6435 \, a^{7} 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]

2/6435*I*(429*(I*a*tan(d*x + c) + a)^(15/2) - 2970*(I*a*tan(d*x + c) + a)^(13/2)*a + 7020*(I*a*tan(d*x + c) +
a)^(11/2)*a^2 - 5720*(I*a*tan(d*x + c) + a)^(9/2)*a^3)/(a^7*d)

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Fricas [A]
time = 0.42, size = 154, normalized size = 1.32 \begin {gather*} -\frac {256 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (15 i \, d x + 15 i \, c\right )} + 120 i \, e^{\left (13 i \, d x + 13 i \, c\right )} + 390 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 715 i \, e^{\left (9 i \, d x + 9 i \, c\right )}\right )}}{6435 \, {\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 {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]

-256/6435*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(16*I*e^(15*I*d*x + 15*I*c) + 120*I*e^(13*I*d*x + 13*I*c)
+ 390*I*e^(11*I*d*x + 11*I*c) + 715*I*e^(9*I*d*x + 9*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sec ^{8}{\left (c + d x \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(sqrt(I*a*(tan(c + d*x) - I))*sec(c + d*x)**8, 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(sqrt(I*a*tan(d*x + c) + a)*sec(d*x + c)^8, x)

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Mupad [B]
time = 12.12, size = 474, normalized size = 4.05 \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}}\,4096{}\mathrm {i}}{6435\,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}}\,2048{}\mathrm {i}}{6435\,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}}\,512{}\mathrm {i}}{2145\,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}}\,256{}\mathrm {i}}{1287\,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}}\,40960{}\mathrm {i}}{1287\,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}}\,52736{}\mathrm {i}}{715\,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}}\,11776{}\mathrm {i}}{195\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\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}}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*40960i)/(1287*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)*2048i)/(6435*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)*512i)/(2145*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)*256i)/(1287
*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)*409
6i)/(6435*d) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*52736i)/(715*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)*11776i)/(195
*d*(exp(c*2i + d*x*2i) + 1)^6) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*256
i)/(15*d*(exp(c*2i + d*x*2i) + 1)^7)

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