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*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))

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Rubi [A]  time = 0.311941, antiderivative size = 169, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.645988, 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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Maple [A]  time = 0.477, size = 132, normalized size = 0.8 \begin{align*} -{\frac{2\,{a}^{2} \left ( 14\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}-14\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-6\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -8\,i\cos \left ( dx+c \right ) -16\,\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{77\,d{e}^{11}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{11}{2}}}} \end{align*}

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*a^2*(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

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Maxima [A]  time = 1.91745, size = 167, normalized size = 0.99 \begin{align*} \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}}} \end{align*}

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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Fricas [A]  time = 2.34185, size = 288, normalized size = 1.7 \begin{align*} \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}} \end{align*}

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)