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]

2/7*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a^2
/d/(e*cos(d*x+c))^(1/2)+2/7*I/d/(e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^2+2/7*I/d/(e*cos(d*x+c))^(1/2)/(a^2+I*
a^2*tan(d*x+c))

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Rubi [A]  time = 0.16, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.61, 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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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \frac {{\left (7 \, a^{2} d e e^{\left (3 i \, d x + 3 i \, c\right )} {\rm integral}\left (-\frac {2 i \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{7 \, {\left (a^{2} d e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e\right )}}, x\right ) + \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{7 \, a^{2} d e} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \cos \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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)

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maple [A]  time = 6.02, size = 240, normalized size = 2.00 \[ \frac {\frac {64 i \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {128 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {96 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {32 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{7}+\frac {12 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{7}+\frac {4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{7}}{a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{\sqrt {e \cos {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )} - 2 i \sqrt {e \cos {\left (c + d x \right )}} \tan {\left (c + d x \right )} - \sqrt {e \cos {\left (c + d x \right )}}}\, dx}{a^{2}} \]

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]

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

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