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

Optimal. Leaf size=89 \[ \frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}-\frac {2 a \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d (e \cos (c+d x))^{3/2}}+\frac {2 a \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

2/3*I*a/d/(e*cos(d*x+c))^(3/2)-2*a*cos(d*x+c)^(3/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(
sin(1/2*d*x+1/2*c),2^(1/2))/d/(e*cos(d*x+c))^(3/2)+2*a*sin(d*x+c)/d/e/(e*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 92, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3515, 3486, 3768, 3771, 2639} \[ \frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}-\frac {2 a \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d (e \cos (c+d x))^{3/2}}+\frac {2 a \sin (c+d x) \cos (c+d x)}{d (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/3)*a)/(d*(e*Cos[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2])/(d*(e*Cos[c + d*
x])^(3/2)) + (2*a*Cos[c + d*x]*Sin[c + d*x])/(d*(e*Cos[c + d*x])^(3/2))

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3486

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

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx &=\frac {\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\\ &=\frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}+\frac {a \int (e \sec (c+d x))^{3/2} \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\\ &=\frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{d (e \cos (c+d x))^{3/2}}-\frac {\left (a e^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\\ &=\frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{d (e \cos (c+d x))^{3/2}}-\frac {\left (a \cos ^{\frac {3}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \, dx}{(e \cos (c+d x))^{3/2}}\\ &=\frac {2 i a}{3 d (e \cos (c+d x))^{3/2}}-\frac {2 a \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d (e \cos (c+d x))^{3/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{d (e \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 3.79, size = 369, normalized size = 4.15 \[ \frac {\cos ^{\frac {5}{2}}(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) \left (\frac {2 \sin (c) (\cot (c)-i) (3 \csc (c) \cos (c+2 d x)+3 \cot (c)+2 i)}{3 \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 \sqrt {2} (\cot (c)-i) e^{-i d x} \left (e^{2 i d x} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+3 e^{2 i (c+d x)}-3 \sqrt {1-i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} F\left (\left .\sin ^{-1}\left (\sqrt {\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3 \sqrt {1-i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} E\left (\left .\sin ^{-1}\left (\sqrt {\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3\right )}{3 \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{2 d (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(5/2)*((-2*Sqrt[2]*(-I + Cot[c])*(3 + 3*E^((2*I)*(c + d*x)) + 3*Sqrt[1 - I*E^(I*(c + d*x))]*Sqrt
[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] - 3*Sqr
t[1 - I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticF[ArcSin[Sqrt[(-I)*Cos[c + d*x]
+ Sin[c + d*x]]], -1] + E^((2*I)*d*x)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)
*(c + d*x))]))/(3*E^(I*d*x)*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (2*(-I + Cot[c])*(2*I + 3*Cot[c
] + 3*Cos[c + 2*d*x]*Csc[c])*Sin[c])/(3*Cos[c + d*x]^(3/2)))*(Cos[d*x] - I*Sin[d*x])*(a + I*a*Tan[c + d*x]))/(
2*d*(e*Cos[c + d*x])^(3/2))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \frac {\sqrt {\frac {1}{2}} {\left (-12 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 3 \, {\left (d e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )} {\rm integral}\left (\frac {2 i \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} a e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}}, x\right )}{3 \, {\left (d e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(1/2)*(-12*I*a*e^(4*I*d*x + 4*I*c) - 4*I*a*e^(2*I*d*x + 2*I*c))*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e^(-1
/2*I*d*x - 1/2*I*c) + 3*(d*e^2*e^(4*I*d*x + 4*I*c) + 2*d*e^2*e^(2*I*d*x + 2*I*c) + d*e^2)*integral(2*I*sqrt(1/
2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*a*e^(1/2*I*d*x + 1/2*I*c)/(d*e^2*e^(2*I*d*x + 2*I*c) + d*e^2), x))/(d*e^2*e
^(4*I*d*x + 4*I*c) + 2*d*e^2*e^(2*I*d*x + 2*I*c) + d*e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 7.90, size = 214, normalized size = 2.40 \[ -\frac {2 \left (6 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e*(6*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-12*sin(
1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(
cos(1/2*d*x+1/2*c),2^(1/2))+6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+I*sin(1/2*d*x+1/2*c))*a/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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