3.415 \(\int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx\)

Optimal. Leaf size=169 \[ \frac {32 i a^3 \sqrt {e \sec (c+d x)}}{77 d e^6 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}} \]

[Out]

32/77*I*a^3*(e*sec(d*x+c))^(1/2)/d/e^6/(a+I*a*tan(d*x+c))^(1/2)-16/77*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/e^4/(e*
sec(d*x+c))^(3/2)-12/77*I*a*(a+I*a*tan(d*x+c))^(3/2)/d/e^2/(e*sec(d*x+c))^(7/2)-2/11*I*(a+I*a*tan(d*x+c))^(5/2
)/d/(e*sec(d*x+c))^(11/2)

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Rubi [A]  time = 0.31, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3497, 3488} \[ \frac {32 i a^3 \sqrt {e \sec (c+d x)}}{77 d e^6 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(11/2),x]

[Out]

(((32*I)/77)*a^3*Sqrt[e*Sec[c + d*x]])/(d*e^6*Sqrt[a + I*a*Tan[c + d*x]]) - (((16*I)/77)*a^2*Sqrt[a + I*a*Tan[
c + d*x]])/(d*e^4*(e*Sec[c + d*x])^(3/2)) - (((12*I)/77)*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*e^2*(e*Sec[c + d*x
])^(7/2)) - (((2*I)/11)*(a + I*a*Tan[c + d*x])^(5/2))/(d*(e*Sec[c + d*x])^(11/2))

Rule 3488

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rule 3497

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d*
Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx &=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}+\frac {(6 a) \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx}{11 e^2}\\ &=-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}+\frac {\left (24 a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{77 e^4}\\ &=-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}+\frac {\left (16 a^3\right ) \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{77 e^6}\\ &=\frac {32 i a^3 \sqrt {e \sec (c+d x)}}{77 d e^6 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 121, normalized size = 0.72 \[ \frac {a^2 \sqrt {a+i a \tan (c+d x)} (-22 \sin (c+d x)+42 \sin (3 (c+d x))-55 i \cos (c+d x)+35 i \cos (3 (c+d x))) (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x)))}{154 d e^5 (\cos (d x)+i \sin (d x))^2 \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(11/2),x]

[Out]

(a^2*((-55*I)*Cos[c + d*x] + (35*I)*Cos[3*(c + d*x)] - 22*Sin[c + d*x] + 42*Sin[3*(c + d*x)])*(Cos[2*(c + 2*d*
x)] + I*Sin[2*(c + 2*d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(154*d*e^5*Sqrt[e*Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x]
)^2)

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fricas [A]  time = 0.45, size = 99, normalized size = 0.59 \[ \frac {{\left (-7 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 40 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 110 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 77 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{308 \, d e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

1/308*(-7*I*a^2*e^(8*I*d*x + 8*I*c) - 40*I*a^2*e^(6*I*d*x + 6*I*c) - 110*I*a^2*e^(4*I*d*x + 4*I*c) + 77*I*a^2)
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/(d*e^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)/(e*sec(d*x + c))^(11/2), x)

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maple [A]  time = 1.42, size = 132, normalized size = 0.78 \[ -\frac {2 \left (14 i \left (\cos ^{5}\left (d x +c \right )\right )-14 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-i \left (\cos ^{3}\left (d x +c \right )\right )-6 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 i \cos \left (d x +c \right )-16 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} \left (\cos ^{6}\left (d x +c \right )\right ) a^{2}}{77 d \,e^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(11/2),x)

[Out]

-2/77/d*(14*I*cos(d*x+c)^5-14*sin(d*x+c)*cos(d*x+c)^4-I*cos(d*x+c)^3-6*cos(d*x+c)^2*sin(d*x+c)-8*I*cos(d*x+c)-
16*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e/cos(d*x+c))^(11/2)*cos(d*x+c)^6/e^11*a^2

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maxima [A]  time = 1.11, size = 124, normalized size = 0.73 \[ \frac {{\left (-7 i \, a^{2} \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) - 33 i \, a^{2} \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 77 i \, a^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 77 i \, a^{2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{2} \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 \, a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 77 \, a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 77 \, a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{308 \, d e^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

1/308*(-7*I*a^2*cos(11/2*d*x + 11/2*c) - 33*I*a^2*cos(7/2*d*x + 7/2*c) - 77*I*a^2*cos(3/2*d*x + 3/2*c) + 77*I*
a^2*cos(1/2*d*x + 1/2*c) + 7*a^2*sin(11/2*d*x + 11/2*c) + 33*a^2*sin(7/2*d*x + 7/2*c) + 77*a^2*sin(3/2*d*x + 3
/2*c) + 77*a^2*sin(1/2*d*x + 1/2*c))*sqrt(a)/(d*e^(11/2))

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mupad [B]  time = 6.07, size = 133, normalized size = 0.79 \[ -\frac {a^2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-187\,\sin \left (2\,c+2\,d\,x\right )-40\,\sin \left (4\,c+4\,d\,x\right )-7\,\sin \left (6\,c+6\,d\,x\right )+\cos \left (2\,c+2\,d\,x\right )\,33{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,40{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,7{}\mathrm {i}\right )}{616\,d\,e^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/(e/cos(c + d*x))^(11/2),x)

[Out]

-(a^2*(e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(
cos(2*c + 2*d*x)*33i + cos(4*c + 4*d*x)*40i + cos(6*c + 6*d*x)*7i - 187*sin(2*c + 2*d*x) - 40*sin(4*c + 4*d*x)
- 7*sin(6*c + 6*d*x)))/(616*d*e^6)

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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((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(11/2),x)

[Out]

Timed out

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