3.407 \(\int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac {16 i a^2}{45 d e^4 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {32 i a \sqrt {a+i a \tan (c+d x)}}{45 d e^4 \sqrt {e \sec (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}} \]

[Out]

16/45*I*a^2/d/e^4/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-4/15*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*sec
(d*x+c))^(5/2)-32/45*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/e^4/(e*sec(d*x+c))^(1/2)-2/9*I*(a+I*a*tan(d*x+c))^(3/2)/d/
(e*sec(d*x+c))^(9/2)

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Rubi [A]  time = 0.29, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3497, 3502, 3488} \[ \frac {16 i a^2}{45 d e^4 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {32 i a \sqrt {a+i a \tan (c+d x)}}{45 d e^4 \sqrt {e \sec (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)/(e*Sec[c + d*x])^(9/2),x]

[Out]

(((16*I)/45)*a^2)/(d*e^4*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/15)*a*Sqrt[a + I*a*Tan[c +
d*x]])/(d*e^2*(e*Sec[c + d*x])^(5/2)) - (((32*I)/45)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^4*Sqrt[e*Sec[c + d*x]
]) - (((2*I)/9)*(a + I*a*Tan[c + d*x])^(3/2))/(d*(e*Sec[c + d*x])^(9/2))

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]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && 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))^{9/2}} \, dx &=-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}+\frac {(2 a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx}{3 e^2}\\ &=-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}+\frac {\left (8 a^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{15 e^4}\\ &=\frac {16 i a^2}{45 d e^4 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}+\frac {(16 a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{45 e^4}\\ &=\frac {16 i a^2}{45 d e^4 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac {32 i a \sqrt {a+i a \tan (c+d x)}}{45 d e^4 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 113, normalized size = 0.68 \[ \frac {a (\cos (d x)-i \sin (d x)) \sqrt {a+i a \tan (c+d x)} (-54 \sin (c+d x)+10 \sin (3 (c+d x))-81 i \cos (c+d x)+5 i \cos (3 (c+d x))) (\cos (c+2 d x)+i \sin (c+2 d x))}{90 d e^4 \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/(e*Sec[c + d*x])^(9/2),x]

[Out]

(a*(Cos[d*x] - I*Sin[d*x])*((-81*I)*Cos[c + d*x] + (5*I)*Cos[3*(c + d*x)] - 54*Sin[c + d*x] + 10*Sin[3*(c + d*
x)])*(Cos[c + 2*d*x] + I*Sin[c + 2*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(90*d*e^4*Sqrt[e*Sec[c + d*x]])

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fricas [A]  time = 1.02, size = 103, normalized size = 0.62 \[ \frac {{\left (-5 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 32 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 162 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 120 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{180 \, d e^{5}} \]

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))^(9/2),x, algorithm="fricas")

[Out]

1/180*(-5*I*a*e^(8*I*d*x + 8*I*c) - 32*I*a*e^(6*I*d*x + 6*I*c) - 162*I*a*e^(4*I*d*x + 4*I*c) - 120*I*a*e^(2*I*
d*x + 2*I*c) + 15*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^(-3/2*I*d*x - 3/2
*I*c)/(d*e^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}}\,{d x} \]

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))^(9/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)/(e*sec(d*x + c))^(9/2), x)

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maple [A]  time = 1.33, size = 113, normalized size = 0.68 \[ -\frac {2 \left (5 i \left (\cos ^{4}\left (d x +c \right )\right )-5 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 i \left (\cos ^{2}\left (d x +c \right )\right )-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )+16 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \left (\cos ^{5}\left (d x +c \right )\right ) a}{45 d \,e^{9}} \]

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))^(9/2),x)

[Out]

-2/45/d*(5*I*cos(d*x+c)^4-5*cos(d*x+c)^3*sin(d*x+c)-2*I*cos(d*x+c)^2-8*cos(d*x+c)*sin(d*x+c)+16*I)*(a*(I*sin(d
*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e/cos(d*x+c))^(9/2)*cos(d*x+c)^5/e^9*a

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maxima [A]  time = 1.38, size = 160, normalized size = 0.96 \[ \frac {{\left (-5 i \, a \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 15 i \, a \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 27 i \, a \cos \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 135 i \, a \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 5 \, a \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 15 \, a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 27 \, a \sin \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 135 \, a \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a}}{180 \, d e^{\frac {9}{2}}} \]

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))^(9/2),x, algorithm="maxima")

[Out]

1/180*(-5*I*a*cos(9/2*d*x + 9/2*c) + 15*I*a*cos(3/2*d*x + 3/2*c) - 27*I*a*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c)
, cos(3/2*d*x + 3/2*c))) - 135*I*a*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 5*a*sin(9/2*
d*x + 9/2*c) + 15*a*sin(3/2*d*x + 3/2*c) + 27*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) +
135*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sqrt(a)/(d*e^(9/2))

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mupad [B]  time = 5.58, size = 125, normalized size = 0.75 \[ -\frac {a\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-42\,\sin \left (c+d\,x\right )-47\,\sin \left (3\,c+3\,d\,x\right )-5\,\sin \left (5\,c+5\,d\,x\right )+\cos \left (c+d\,x\right )\,282{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,17{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,5{}\mathrm {i}\right )}{360\,d\,e^5} \]

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))^(9/2),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)*(co
s(c + d*x)*282i - 42*sin(c + d*x) + cos(3*c + 3*d*x)*17i + cos(5*c + 5*d*x)*5i - 47*sin(3*c + 3*d*x) - 5*sin(5
*c + 5*d*x)))/(360*d*e^5)

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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))**(9/2),x)

[Out]

Timed out

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