3.5.28 \(\int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \, dx\) [428]

Optimal. Leaf size=165 \[ \frac {2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i \sqrt {e \sec (c+d x)}}{45 a d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 165, 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 = {3583, 3578, 3569} \begin {gather*} -\frac {16 i \sqrt {a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}+\frac {32 i \sqrt {e \sec (c+d x)}}{45 a d e^2 \sqrt {a+i a \tan (c+d x)}}+\frac {4 i}{15 a d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

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

Rule 3569

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 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[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 3583

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 {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {2 \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx}{3 a}\\ &=\frac {2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{15 a^2}\\ &=\frac {2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}+\frac {16 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{45 a e^2}\\ &=\frac {2 i}{9 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {4 i}{15 a d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i \sqrt {e \sec (c+d x)}}{45 a d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{45 a^2 d (e \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 100, normalized size = 0.61 \begin {gather*} -\frac {\sec ^3(c+d x) (-81 \cos (c+d x)+5 \cos (3 (c+d x))-54 i \sin (c+d x)+10 i \sin (3 (c+d x)))}{90 a d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

-1/90*(Sec[c + d*x]^3*(-81*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - (54*I)*Sin[c + d*x] + (10*I)*Sin[3*(c + d*x)]))
/(a*d*(e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.89, size = 132, normalized size = 0.80

method	result	size
default	\(\frac {2 \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (10 i \left (\cos ^{5}\left (d x +c \right )\right )+10 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+i \left (\cos ^{3}\left (d x +c \right )\right )+6 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+8 i \cos \left (d x +c \right )+16 \sin \left (d x +c \right )\right )}{45 d \,e^{3} a^{2}}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/45/d*cos(d*x+c)^2*(e/cos(d*x+c))^(3/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(10*I*cos(d*x+c)^5+10*
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))/e^3/a^2

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Maxima [A]
time = 0.55, size = 177, normalized size = 1.07 \begin {gather*} \frac {{\left (5 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 27 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) - 15 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 135 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 27 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 15 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 135 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )\right )} e^{\left (-\frac {3}{2}\right )}}{180 \, a^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 94, normalized size = 0.57 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-15 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 120 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 162 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 32 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c - \frac {3}{2}\right )}}{180 \, a^{2} d \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/180*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-15*I*e^(8*I*d*x + 8*I*c) + 120*I*e^(6*I*d*x + 6*I*c) + 162*I*e^(4*I*
d*x + 4*I*c) + 32*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-9/2*I*d*x - 9/2*I*c - 3/2)/(a^2*d*sqrt(e^(2*I*d*x + 2*I*c)
+ 1))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(1/((e*sec(c + d*x))**(3/2)*(I*a*(tan(c + d*x) - I))**(3/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.20, size = 112, normalized size = 0.68 \begin {gather*} \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,12{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}+42\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (4\,c+4\,d\,x\right )+135{}\mathrm {i}\right )}{180\,a\,d\,e^2\,\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*12i + cos(4*c + 4*d*x)*5i + 42*sin(2*c + 2*d*x) + 5*sin(4*c + 4*d*x)
+ 135i))/(180*a*d*e^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))

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