3.3.32 \(\int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))} \, dx\) [232]

Optimal. Leaf size=145 \[ \frac {30 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{77 a d e^4}+\frac {18 \sin (c+d x)}{77 a d e (e \sec (c+d x))^{5/2}}+\frac {30 \sin (c+d x)}{77 a d e^3 \sqrt {e \sec (c+d x)}}+\frac {2 i}{11 d (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))} \]

[Out]

18/77*sin(d*x+c)/a/d/e/(e*sec(d*x+c))^(5/2)+30/77*sin(d*x+c)/a/d/e^3/(e*sec(d*x+c))^(1/2)+30/77*(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)*(e*sec(d*x+c))^(1/2)/
a/d/e^4+2/11*I/d/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))

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Rubi [A]
time = 0.09, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3583, 3854, 3856, 2720} \begin {gather*} \frac {30 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{77 a d e^4}+\frac {30 \sin (c+d x)}{77 a d e^3 \sqrt {e \sec (c+d x)}}+\frac {18 \sin (c+d x)}{77 a d e (e \sec (c+d x))^{5/2}}+\frac {2 i}{11 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

(30*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*a*d*e^4) + (18*Sin[c + d*x])/(77*a*
d*e*(e*Sec[c + d*x])^(5/2)) + (30*Sin[c + d*x])/(77*a*d*e^3*Sqrt[e*Sec[c + d*x]]) + ((2*I)/11)/(d*(e*Sec[c + d
*x])^(7/2)*(a + I*a*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 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 {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))} \, dx &=\frac {2 i}{11 d (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))}+\frac {9 \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{11 a}\\ &=\frac {18 \sin (c+d x)}{77 a d e (e \sec (c+d x))^{5/2}}+\frac {2 i}{11 d (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))}+\frac {45 \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{77 a e^2}\\ &=\frac {18 \sin (c+d x)}{77 a d e (e \sec (c+d x))^{5/2}}+\frac {30 \sin (c+d x)}{77 a d e^3 \sqrt {e \sec (c+d x)}}+\frac {2 i}{11 d (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))}+\frac {15 \int \sqrt {e \sec (c+d x)} \, dx}{77 a e^4}\\ &=\frac {18 \sin (c+d x)}{77 a d e (e \sec (c+d x))^{5/2}}+\frac {30 \sin (c+d x)}{77 a d e^3 \sqrt {e \sec (c+d x)}}+\frac {2 i}{11 d (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))}+\frac {\left (15 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 a e^4}\\ &=\frac {30 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{77 a d e^4}+\frac {18 \sin (c+d x)}{77 a d e (e \sec (c+d x))^{5/2}}+\frac {30 \sin (c+d x)}{77 a d e^3 \sqrt {e \sec (c+d x)}}+\frac {2 i}{11 d (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 142, normalized size = 0.98 \begin {gather*} -\frac {(e \sec (c+d x))^{3/2} \left (-148 \cos (c+d x)+34 \cos (3 (c+d x))+2 \cos (5 (c+d x))+240 i \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))+78 i \sin (c+d x)+87 i \sin (3 (c+d x))+9 i \sin (5 (c+d x))\right )}{616 a d e^5 (-i+\tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

-1/616*((e*Sec[c + d*x])^(3/2)*(-148*Cos[c + d*x] + 34*Cos[3*(c + d*x)] + 2*Cos[5*(c + d*x)] + (240*I)*Sqrt[Co
s[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] + I*Sin[c + d*x]) + (78*I)*Sin[c + d*x] + (87*I)*Sin[3*(c
+ d*x)] + (9*I)*Sin[5*(c + d*x)]))/(a*d*e^5*(-I + Tan[c + d*x]))

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Maple [A]
time = 1.44, size = 236, normalized size = 1.63

method	result	size
default	\(\frac {2 \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (7 i \left (\cos ^{6}\left (d x +c \right )\right )+7 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+9 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+15 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{77 a d \,e^{7} \sin \left (d x +c \right )^{4}}\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/77/a/d*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*cos(d*x+c)^3*(e/cos(d*x+c))^(7/2)*(7*I*cos(d*x+c)^6+7*sin(d*x+c)*c
os(d*x+c)^5+15*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)+15*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)+9*sin(d*x+c)*cos(d*x+c)^3+15*sin(d*x+c)*cos(d*x+c))/e^7/sin(d*x+c)^4

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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(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 128, normalized size = 0.88 \begin {gather*} \frac {{\left (-480 i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-11 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 121 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 70 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 226 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 53 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\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 (-6 i \, d x - 6 i \, c - \frac {7}{2}\right )}}{1232 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/1232*(-480*I*sqrt(2)*e^(6*I*d*x + 6*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-11*I*e^(10*
I*d*x + 10*I*c) - 121*I*e^(8*I*d*x + 8*I*c) + 70*I*e^(6*I*d*x + 6*I*c) + 226*I*e^(4*I*d*x + 4*I*c) + 53*I*e^(2
*I*d*x + 2*I*c) + 7*I)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-6*I*d*x - 6*I*c - 7/2)/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}} \tan {\left (c + d x \right )} - i \left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c)),x)

[Out]

-I*Integral(1/((e*sec(c + d*x))**(7/2)*tan(c + d*x) - I*(e*sec(c + d*x))**(7/2)), x)/a

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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(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(e^(-7/2)/((I*a*tan(d*x + c) + a)*sec(d*x + c)^(7/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)),x)

[Out]

int(1/((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)), x)

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