3.262 \(\int \frac{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=191 \[ \frac{4 i e^2}{33 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}+\frac{2 e \sin (c+d x)}{33 a^4 d \sqrt{e \sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 a^4 d}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4} \]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*a^4*d) + (2*e*Sin[c + d*x])/(33*a^4*
d*Sqrt[e*Sec[c + d*x]]) + (((2*I)/15)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^4) + (((14*I)/165)*Sqrt[
e*Sec[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^3) + (((4*I)/33)*e^2)/(d*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[
c + d*x]))

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Rubi [A]  time = 0.191723, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3502, 3500, 3769, 3771, 2641} \[ \frac{4 i e^2}{33 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}+\frac{2 e \sin (c+d x)}{33 a^4 d \sqrt{e \sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 a^4 d}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sec[c + d*x]]/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*a^4*d) + (2*e*Sin[c + d*x])/(33*a^4*
d*Sqrt[e*Sec[c + d*x]]) + (((2*I)/15)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^4) + (((14*I)/165)*Sqrt[
e*Sec[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^3) + (((4*I)/33)*e^2)/(d*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[
c + d*x]))

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]

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{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx &=\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac{7 \int \frac{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx}{15 a}\\ &=\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{7 \int \frac{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{33 a^2}\\ &=\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{e^2 \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^4}\\ &=\frac{2 e \sin (c+d x)}{33 a^4 d \sqrt{e \sec (c+d x)}}+\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\int \sqrt{e \sec (c+d x)} \, dx}{33 a^4}\\ &=\frac{2 e \sin (c+d x)}{33 a^4 d \sqrt{e \sec (c+d x)}}+\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{33 a^4}\\ &=\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 a^4 d}+\frac{2 e \sin (c+d x)}{33 a^4 d \sqrt{e \sec (c+d x)}}+\frac{2 i \sqrt{e \sec (c+d x)}}{15 d (a+i a \tan (c+d x))^4}+\frac{14 i \sqrt{e \sec (c+d x)}}{165 a d (a+i a \tan (c+d x))^3}+\frac{4 i e^2}{33 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.486795, size = 137, normalized size = 0.72 \[ \frac{\sec ^4(c+d x) \sqrt{e \sec (c+d x)} \left (i (54 i \sin (2 (c+d x))+37 i \sin (4 (c+d x))+112 \cos (2 (c+d x))+48 \cos (4 (c+d x))+64)+40 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (4 (c+d x))+i \sin (4 (c+d x)))\right )}{660 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Sec[c + d*x]]/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[4*(c + d*x)] + I*Si
n[4*(c + d*x)]) + I*(64 + 112*Cos[2*(c + d*x)] + 48*Cos[4*(c + d*x)] + (54*I)*Sin[2*(c + d*x)] + (37*I)*Sin[4*
(c + d*x)])))/(660*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.356, size = 252, normalized size = 1.3 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{165\,{a}^{4}d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}} \left ( 88\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+88\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}-60\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-16\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +5\,i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^4,x)

[Out]

2/165/a^4/d*(e/cos(d*x+c))^(1/2)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*(88*I*cos(d*x+c)^8+88*sin(d*x+c)*cos(d*x+c)
^7-60*I*cos(d*x+c)^6-16*cos(d*x+c)^5*sin(d*x+c)+5*I*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+3*cos(d*x+c)^3*sin(d*x+c)+5*cos(d*x+c)*sin(d*x+c))/sin(d*x+c)^4

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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((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (1320 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}{\rm integral}\left (-\frac{i \, \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}}{33 \, a^{4} d}, x\right ) + \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (85 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 166 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 128 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 58 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 11 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{1320 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/1320*(1320*a^4*d*e^(8*I*d*x + 8*I*c)*integral(-1/33*I*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*
x - 1/2*I*c)/(a^4*d), x) + sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(85*I*e^(8*I*d*x + 8*I*c) + 166*I*e^(6*I*
d*x + 6*I*c) + 128*I*e^(4*I*d*x + 4*I*c) + 58*I*e^(2*I*d*x + 2*I*c) + 11*I)*e^(1/2*I*d*x + 1/2*I*c))*e^(-8*I*d
*x - 8*I*c)/(a^4*d)

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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((e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e \sec \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(sqrt(e*sec(d*x + c))/(I*a*tan(d*x + c) + a)^4, x)