3.3.62 \(\int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^4} \, dx\) [262]

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

[Out]

2/33*e*sin(d*x+c)/a^4/d/(e*sec(d*x+c))^(1/2)+2/33*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/a^4/d+2/15*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan
(d*x+c))^4+14/165*I*(e*sec(d*x+c))^(1/2)/a/d/(a+I*a*tan(d*x+c))^3+4/33*I*e^2/d/(e*sec(d*x+c))^(3/2)/(a^4+I*a^4
*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 191, normalized size of antiderivative = 1.00, 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 = {3583, 3581, 3854, 3856, 2720} \begin {gather*} \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} \end {gather*}

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 2720

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

Rule 3581

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 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]

Rule 3854

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 3856

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 {\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 = 0.68, size = 137, normalized size = 0.72 \begin {gather*} \frac {\sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (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)))+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)))\right )}{660 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 0.76, size = 224, normalized size = 1.17

method	result	size
default	\(\frac {2 \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \left (88 i \left (\cos ^{8}\left (d x +c \right )\right )+88 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-60 i \left (\cos ^{6}\left (d x +c \right )\right )-16 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+5 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{165 a^{4} d}\)	\(224\)

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,method=_RETURNVERBOSE)

[Out]

2/165/a^4/d*(e/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*sin(d*x+c)
*cos(d*x+c)^5+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*EllipticF(I*(-1+cos(d*
x+c))/sin(d*x+c),I)+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))
/sin(d*x+c),I)+3*sin(d*x+c)*cos(d*x+c)^3+5*sin(d*x+c)*cos(d*x+c))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 124, normalized size = 0.65 \begin {gather*} \frac {{\left (-80 i \, \sqrt {2} e^{\left (8 i \, d x + 8 i \, c + \frac {1}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (11 i \, e^{\frac {1}{2}} + 85 i \, e^{\left (8 i \, d x + 8 i \, c + \frac {1}{2}\right )} + 166 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {1}{2}\right )} + 128 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} + 58 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{1320 \, a^{4} d} \end {gather*}

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*(-80*I*sqrt(2)*e^(8*I*d*x + 8*I*c + 1/2)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(11*I*e^
(1/2) + 85*I*e^(8*I*d*x + 8*I*c + 1/2) + 166*I*e^(6*I*d*x + 6*I*c + 1/2) + 128*I*e^(4*I*d*x + 4*I*c + 1/2) + 5
8*I*e^(2*I*d*x + 2*I*c + 1/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-8*I*d*x - 8*I*c)/(a^
4*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

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]

Integral(sqrt(e*sec(c + d*x))/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) +
1), x)/a**4

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________