3.667 \(\int \frac{1}{\sqrt{e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=120 \[ \frac{2 i}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt{e \cos (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{2 i}{7 d (a+i a \tan (c+d x))^2 \sqrt{e \cos (c+d x)}} \]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + ((2*I)/7)/(d*Sqrt[e*Cos[c +
d*x]]*(a + I*a*Tan[c + d*x])^2) + ((2*I)/7)/(d*Sqrt[e*Cos[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.158404, antiderivative size = 126, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3515, 3500, 3769, 3771, 2641} \[ \frac{4 i \cos ^2(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt{e \cos (c+d x)}}+\frac{2 \sin (c+d x) \cos (c+d x)}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 a^2 d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (2*Cos[c + d*x]*Sin[c + d*x]
)/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (((4*I)/7)*Cos[c + d*x]^2)/(d*Sqrt[e*Cos[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x
]))

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 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3769

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

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx &=\frac{\int \frac{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{\sqrt{e \cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{4 i \cos ^2(c+d x)}{7 d \sqrt{e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 e^2\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a^2 \sqrt{e \cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 \cos (c+d x) \sin (c+d x)}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 i \cos ^2(c+d x)}{7 d \sqrt{e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\int \sqrt{e \sec (c+d x)} \, dx}{7 a^2 \sqrt{e \cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 \cos (c+d x) \sin (c+d x)}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 i \cos ^2(c+d x)}{7 d \sqrt{e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\sqrt{\cos (c+d x)} \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{7 a^2 \sqrt{e \cos (c+d x)}}\\ &=\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{2 \cos (c+d x) \sin (c+d x)}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 i \cos ^2(c+d x)}{7 d \sqrt{e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.640359, size = 158, normalized size = 1.32 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )-i \cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sqrt{\cos (c+d x)} \left (4 i \sin ^3\left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{3}{2} (c+d x)\right )\right )+2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (\sin \left (\frac{3}{2} (c+d x)\right )-i \cos \left (\frac{3}{2} (c+d x)\right )\right )\right )}{7 a^2 d \cos ^{\frac{3}{2}}(c+d x) (\tan (c+d x)-i)^2 \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Sqrt[Cos[c + d*x]]*(3*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] + (
4*I)*Sin[(c + d*x)/2]^3) + 2*EllipticF[(c + d*x)/2, 2]*((-I)*Cos[(3*(c + d*x))/2] + Sin[(3*(c + d*x))/2])))/(7
*a^2*d*Cos[c + d*x]^(3/2)*Sqrt[e*Cos[c + d*x]]*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 2.564, size = 240, normalized size = 2. \begin{align*}{\frac{2}{7\,{a}^{2}d} \left ( 32\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}-32\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-64\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}+48\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +48\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}-28\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -16\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}-\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +6\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(32*I*sin(1/2*d*x+1/2*c)^9-32*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^8-64*I*sin(1/2*d*x+1/2*c)^7+48*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+48*I*sin(1/2*d*x+1
/2*c)^5-28*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-16*I*sin(1/2*d*x+1/2*c)^3-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(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)+2
*I*sin(1/2*d*x+1/2*c))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (7 \, a^{2} d e e^{\left (3 i \, d x + 3 i \, c\right )}{\rm integral}\left (-\frac{2 i \, \sqrt{\frac{1}{2}} \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 )}}{7 \,{\left (a^{2} d e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e\right )}}, x\right ) + \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e}{\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{7 \, a^{2} d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(7*a^2*d*e*e^(3*I*d*x + 3*I*c)*integral(-2/7*I*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e^(-1/2*I*d*x - 1
/2*I*c)/(a^2*d*e*e^(2*I*d*x + 2*I*c) + a^2*d*e), x) + sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(3*I*e^(2*I*d*
x + 2*I*c) + I)*e^(-1/2*I*d*x - 1/2*I*c))*e^(-3*I*d*x - 3*I*c)/(a^2*d*e)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e \cos \left (d x + c\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(e*cos(d*x + c))*(I*a*tan(d*x + c) + a)^2), x)