3.406 \(\int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac {16 i a^2 \sqrt {e \sec (c+d x)}}{21 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

16/21*I*a^2*(e*sec(d*x+c))^(1/2)/d/e^4/(a+I*a*tan(d*x+c))^(1/2)-8/21*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*sec
(d*x+c))^(3/2)-2/7*I*(a+I*a*tan(d*x+c))^(3/2)/d/(e*sec(d*x+c))^(7/2)

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Rubi [A]  time = 0.23, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3497, 3488} \[ \frac {16 i a^2 \sqrt {e \sec (c+d x)}}{21 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)/(e*Sec[c + d*x])^(7/2),x]

[Out]

(((16*I)/21)*a^2*Sqrt[e*Sec[c + d*x]])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((8*I)/21)*a*Sqrt[a + I*a*Tan[c +
d*x]])/(d*e^2*(e*Sec[c + d*x])^(3/2)) - (((2*I)/7)*(a + I*a*Tan[c + d*x])^(3/2))/(d*(e*Sec[c + d*x])^(7/2))

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))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx &=-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}}+\frac {(4 a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{7 e^2}\\ &=-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}}+\frac {\left (8 a^2\right ) \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{21 e^4}\\ &=\frac {16 i a^2 \sqrt {e \sec (c+d x)}}{21 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 98, normalized size = 0.78 \[ \frac {a (\cos (d x)-i \sin (d x)) \sqrt {a+i a \tan (c+d x)} (12 \sin (2 (c+d x))+9 i \cos (2 (c+d x))-7 i) (\cos (c+2 d x)+i \sin (c+2 d x))}{21 d e^3 \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/(e*Sec[c + d*x])^(7/2),x]

[Out]

(a*(Cos[d*x] - I*Sin[d*x])*(-7*I + (9*I)*Cos[2*(c + d*x)] + 12*Sin[2*(c + d*x)])*(Cos[c + 2*d*x] + I*Sin[c + 2
*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*e^3*Sqrt[e*Sec[c + d*x]])

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fricas [A]  time = 0.56, size = 91, normalized size = 0.73 \[ \frac {{\left (-3 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 17 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 21 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{42 \, d e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/42*(-3*I*a*e^(6*I*d*x + 6*I*c) - 17*I*a*e^(4*I*d*x + 4*I*c) + 7*I*a*e^(2*I*d*x + 2*I*c) + 21*I*a)*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^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)/(e*sec(d*x + c))^(7/2), x)

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maple [A]  time = 1.28, size = 103, normalized size = 0.82 \[ -\frac {2 \left (3 i \left (\cos ^{3}\left (d x +c \right )\right )-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 i \cos \left (d x +c \right )-8 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (\cos ^{4}\left (d x +c \right )\right ) a}{21 d \,e^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(7/2),x)

[Out]

-2/21/d*(3*I*cos(d*x+c)^3-3*cos(d*x+c)^2*sin(d*x+c)-4*I*cos(d*x+c)-8*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))^(7/2)*cos(d*x+c)^4/e^7*a

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maxima [A]  time = 1.07, size = 84, normalized size = 0.67 \[ \frac {{\left (-3 i \, a \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 14 i \, a \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 21 i \, a \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 14 \, a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 21 \, a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{42 \, d e^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/42*(-3*I*a*cos(7/2*d*x + 7/2*c) - 14*I*a*cos(3/2*d*x + 3/2*c) + 21*I*a*cos(1/2*d*x + 1/2*c) + 3*a*sin(7/2*d*
x + 7/2*c) + 14*a*sin(3/2*d*x + 3/2*c) + 21*a*sin(1/2*d*x + 1/2*c))*sqrt(a)/(d*e^(7/2))

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mupad [B]  time = 4.92, size = 110, normalized size = 0.88 \[ \frac {a\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,4{}\mathrm {i}-\cos \left (4\,c+4\,d\,x\right )\,3{}\mathrm {i}+38\,\sin \left (2\,c+2\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+7{}\mathrm {i}\right )}{84\,d\,e^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/(e/cos(c + d*x))^(7/2),x)

[Out]

(a*(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)*4i - cos(4*c + 4*d*x)*3i + 38*sin(2*c + 2*d*x) + 3*sin(4*c + 4*d*x) + 7i))/(84*d*e^4)

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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))**(3/2)/(e*sec(d*x+c))**(7/2),x)

[Out]

Timed out

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