3.307 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=88 \[ -\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d} \]

[Out]

-8/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^3/d+8/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^4/d-2/15*I*(a+I*a*tan(d*x+c))^(15/2
)/a^5/d

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Rubi [A]  time = 0.08, 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} \[ -\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

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Mathematica [A]  time = 0.79, size = 97, normalized size = 1.10 \[ \frac {2 a^2 \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} (-187 i \sin (2 (c+d x))+203 \cos (2 (c+d x))+60) (\sin (5 c+7 d x)-i \cos (5 c+7 d x))}{2145 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*a^2*Sec[c + d*x]^7*(60 + 203*Cos[2*(c + d*x)] - (187*I)*Sin[2*(c + d*x)])*((-I)*Cos[5*c + 7*d*x] + Sin[5*c
+ 7*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(2145*d*(Cos[d*x] + I*Sin[d*x])^2)

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fricas [B]  time = 0.68, size = 152, normalized size = 1.73 \[ \frac {\sqrt {2} {\left (-2048 i \, a^{2} e^{\left (15 i \, d x + 15 i \, c\right )} - 15360 i \, a^{2} e^{\left (13 i \, d x + 13 i \, c\right )} - 49920 i \, a^{2} e^{\left (11 i \, d x + 11 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{2145 \, {\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/2145*sqrt(2)*(-2048*I*a^2*e^(15*I*d*x + 15*I*c) - 15360*I*a^2*e^(13*I*d*x + 13*I*c) - 49920*I*a^2*e^(11*I*d*
x + 11*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c)^6, x)

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maple [B]  time = 2.82, size = 144, normalized size = 1.64 \[ -\frac {2 \left (512 i \left (\cos ^{7}\left (d x +c \right )\right )-512 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )+64 i \left (\cos ^{5}\left (d x +c \right )\right )-320 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+28 i \left (\cos ^{3}\left (d x +c \right )\right )-252 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-341 i \cos \left (d x +c \right )+143 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{2145 d \cos \left (d x +c \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-2/2145/d*(512*I*cos(d*x+c)^7-512*sin(d*x+c)*cos(d*x+c)^6+64*I*cos(d*x+c)^5-320*sin(d*x+c)*cos(d*x+c)^4+28*I*c
os(d*x+c)^3-252*cos(d*x+c)^2*sin(d*x+c)-341*I*cos(d*x+c)+143*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*a^2

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maxima [A]  time = 0.45, size = 58, normalized size = 0.66 \[ -\frac {2 i \, {\left (143 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 660 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 780 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2}\right )}}{2145 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/2145*I*(143*(I*a*tan(d*x + c) + a)^(15/2) - 660*(I*a*tan(d*x + c) + a)^(13/2)*a + 780*(I*a*tan(d*x + c) + a
)^(11/2)*a^2)/(a^5*d)

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mupad [B]  time = 11.32, size = 498, normalized size = 5.66 \[ -\frac {a^2\,\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}}{2145\,d}-\frac {a^2\,\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}}{2145\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,\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}}{715\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^2\,\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}}\,18176{}\mathrm {i}}{429\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^2\,\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}}{429\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {a^2\,\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}}\,103936{}\mathrm {i}}{715\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {a^2\,\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}}\,15616{}\mathrm {i}}{195\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}+\frac {a^2\,\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/cos(c + d*x)^6,x)

[Out]

(a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*18176i)/(429*d*(exp(c*2i + d*x*2
i) + 1)^3) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*1024i)/(2145*d*(exp
(c*2i + d*x*2i) + 1)) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*256i)/(7
15*d*(exp(c*2i + d*x*2i) + 1)^2) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/
2)*2048i)/(2145*d) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*52736i)/(42
9*d*(exp(c*2i + d*x*2i) + 1)^4) + (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2
)*103936i)/(715*d*(exp(c*2i + d*x*2i) + 1)^5) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*
2i) + 1))^(1/2)*15616i)/(195*d*(exp(c*2i + d*x*2i) + 1)^6) + (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(ex
p(c*2i + d*x*2i) + 1))^(1/2)*256i)/(15*d*(exp(c*2i + d*x*2i) + 1)^7)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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