3.279 \(\int \sec ^8(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\)

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

[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)

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Rubi [A]  time = 0.0754535, antiderivative size = 117, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{15/2}}{15 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{13/2}}{13 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{11/2}}{11 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{9/2}}{9 a^4 d} \]

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 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 ^8(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x)^{7/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{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*}

Mathematica [A]  time = 0.818673, size = 95, normalized size = 0.81 \[ \frac{2 \sec ^7(c+d x) \sqrt{a+i a \tan (c+d x)} (-3 i (90 \sin (c+d x)+233 \sin (3 (c+d x)))+510 \cos (c+d x)+731 \cos (3 (c+d x))) (\sin (4 (c+d x))-i \cos (4 (c+d x)))}{6435 d} \]

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 = 2.026, size = 141, normalized size = 1.2 \begin{align*}{\frac{-2048\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}+2048\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) -256\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1280\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-112\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1008\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -66\,i\cos \left ( dx+c \right ) +858\,\sin \left ( dx+c \right ) }{6435\,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)^8*(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/6435/d*(-1024*I*cos(d*x+c)^7+1024*cos(d*x+c)^6*sin(d*x+c)-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 = 0.979973, size = 103, normalized size = 0.88 \begin{align*} \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{align*}

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 = 2.37633, size = 522, normalized size = 4.46 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-4096 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 30720 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 99840 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 183040 i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\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{align*}

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]

1/6435*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-4096*I*e^(14*I*d*x + 14*I*c) - 30720*I*e^(12*I*d*x + 12*I*c
) - 99840*I*e^(10*I*d*x + 10*I*c) - 183040*I*e^(8*I*d*x + 8*I*c))*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)**8*(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 \sqrt{i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{8}\,{d x} \end{align*}

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)