3.334 \(\int \frac {\sec ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=88 \[ -\frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, 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 = {3487, 43} \[ -\frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.37, size = 77, normalized size = 0.88 \[ \frac {2 \sec ^5(c+d x) (-55 i \sin (2 (c+d x))+71 \cos (2 (c+d x))+36) (\sin (3 (c+d x))-i \cos (3 (c+d x)))}{315 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]^5*(36 + 71*Cos[2*(c + d*x)] - (55*I)*Sin[2*(c + d*x)])*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x
)]))/(315*d*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.78, size = 113, normalized size = 1.28 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-256 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 1152 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 2016 i \, e^{\left (5 i \, d x + 5 i \, c\right )}\right )}}{315 \, {\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-256*I*e^(9*I*d*x + 9*I*c) - 1152*I*e^(7*I*d*x + 7*I*c) - 201
6*I*e^(5*I*d*x + 5*I*c))/(a*d*e^(8*I*d*x + 8*I*c) + 4*a*d*e^(6*I*d*x + 6*I*c) + 6*a*d*e^(4*I*d*x + 4*I*c) + 4*
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 \[ \int \frac {\sec \left (d x + c\right )^{6}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^6/sqrt(I*a*tan(d*x + c) + a), x)

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maple [A]  time = 1.29, size = 100, normalized size = 1.14 \[ -\frac {2 \left (64 i \left (\cos ^{4}\left (d x +c \right )\right )-64 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+8 i \left (\cos ^{2}\left (d x +c \right )\right )-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )+35 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{315 d \cos \left (d x +c \right )^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

-2/315/d*(64*I*cos(d*x+c)^4-64*cos(d*x+c)^3*sin(d*x+c)+8*I*cos(d*x+c)^2-40*cos(d*x+c)*sin(d*x+c)+35*I)*(a*(I*s
in(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4/a

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maxima [B]  time = 0.45, size = 169, normalized size = 1.92 \[ -\frac {2 i \, {\left (315 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} - \frac {42 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{a^{2}} + \frac {35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 180 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 378 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}}{a^{4}}\right )}}{315 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/315*I*(315*sqrt(I*a*tan(d*x + c) + a) - 42*(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 + (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)/(a*d)

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mupad [B]  time = 6.38, size = 306, normalized size = 3.48 \[ -\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{315\,a\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{315\,a\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{105\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,320{}\mathrm {i}}{63\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(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)*320i)/(63*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)*128i)/(315*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)*32i)/(105*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)*256i)/(315*a*
d) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(9*a*d*(exp(c*2i + d*x*2i)
+ 1)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**6/sqrt(I*a*(tan(c + d*x) - I)), x)

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