3.7.63 \(\int \frac {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx\) [663]

Optimal. Leaf size=190 \[ \frac {2 (e \cos (c+d x))^{7/2} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

2/7*(e*cos(d*x+c))^(7/2)*(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))
/a^2/d/cos(d*x+c)^(7/2)+2/15*cos(d*x+c)*(e*cos(d*x+c))^(7/2)*sin(d*x+c)/a^2/d+6/35*(e*cos(d*x+c))^(7/2)*tan(d*
x+c)/a^2/d+2/7*(e*cos(d*x+c))^(7/2)*sec(d*x+c)^2*tan(d*x+c)/a^2/d+4/15*I*cos(d*x+c)^2*(e*cos(d*x+c))^(7/2)/d/(
a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.17, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3854, 3856, 2720} \begin {gather*} \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{7/2}}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{7/2}}{15 a^2 d}+\frac {6 \tan (c+d x) (e \cos (c+d x))^{7/2}}{35 a^2 d}+\frac {2 \tan (c+d x) \sec ^2(c+d x) (e \cos (c+d x))^{7/2}}{7 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*Cos[c + d*x])^(7/2)*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Cos[c + d*x]^(7/2)) + (2*Cos[c + d*x]*(e*Cos[c +
d*x])^(7/2)*Sin[c + d*x])/(15*a^2*d) + (6*(e*Cos[c + d*x])^(7/2)*Tan[c + d*x])/(35*a^2*d) + (2*(e*Cos[c + d*x
])^(7/2)*Sec[c + d*x]^2*Tan[c + d*x])/(7*a^2*d) + (((4*I)/15)*Cos[c + d*x]^2*(e*Cos[c + d*x])^(7/2))/(d*(a^2 +
I*a^2*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 3596

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 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 {(e \cos (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx &=\left ((e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx\\ &=\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^2 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{11/2}} \, dx}{15 a^2}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{5 a^2}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a^2 e^2}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left ((e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \sqrt {e \sec (c+d x)} \, dx}{7 a^2 e^4}\\ &=\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {(e \cos (c+d x))^{7/2} \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 a^2 \cos ^{\frac {7}{2}}(c+d x)}\\ &=\frac {2 (e \cos (c+d x))^{7/2} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{7/2} \sin (c+d x)}{15 a^2 d}+\frac {6 (e \cos (c+d x))^{7/2} \tan (c+d x)}{35 a^2 d}+\frac {2 (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \tan (c+d x)}{7 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{7/2}}{15 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.36, size = 156, normalized size = 0.82 \begin {gather*} \frac {e^3 \sqrt {e \cos (c+d x)} \left (-240 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\cos (c+d x)} (-296 i \cos (c+d x)+68 i \cos (3 (c+d x))+4 i \cos (5 (c+d x))+134 \sin (c+d x)-117 \sin (3 (c+d x))-11 \sin (5 (c+d x)))\right )}{840 a^2 d \cos ^{\frac {5}{2}}(c+d x) (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[e*Cos[c + d*x]]*(-240*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + Sqrt[Cos[c
+ d*x]]*((-296*I)*Cos[c + d*x] + (68*I)*Cos[3*(c + d*x)] + (4*I)*Cos[5*(c + d*x)] + 134*Sin[c + d*x] - 117*Si
n[3*(c + d*x)] - 11*Sin[5*(c + d*x)])))/(840*a^2*d*Cos[c + d*x]^(5/2)*(-I + Tan[c + d*x])^2)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (192 ) = 384\).
time = 1.91, size = 387, normalized size = 2.04

method	result	size
default	\(-\frac {2 e^{4} \left (-3584 i \left (\sin ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3584 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1568 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12544 \left (\sin ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+14336 i \left (\sin ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19264 \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-25088 i \left (\sin ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16800 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+224 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9104 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-14 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-3128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15680 i \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+700 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6272 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-90 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 \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 )+25088 i \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{105 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}\)	\(387\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^4*(-3584*I*sin(1/2*d*x+1/2*c)^17+3584*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^16-1568*I*sin(1/2*d*x+1/2*c)^5-12544*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c
)+14336*I*sin(1/2*d*x+1/2*c)^15+19264*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)-25088*I*sin(1/2*d*x+1/2*c)^13-1
6800*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+224*I*sin(1/2*d*x+1/2*c)^3+9104*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8-14*I*sin(1/2*d*x+1/2*c)-3128*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-15680*I*sin(1/2*d*x+1/2*c)^9+700*
sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6272*I*sin(1/2*d*x+1/2*c)^7-90*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)
+15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+25088*
I*sin(1/2*d*x+1/2*c)^11)/d

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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((e*cos(d*x+c))^(7/2)/(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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 136, normalized size = 0.72 \begin {gather*} \frac {{\left (\sqrt {\frac {1}{2}} {\left (7 i \, e^{\frac {7}{2}} - 15 i \, e^{\left (10 i \, d x + 10 i \, c + \frac {7}{2}\right )} - 185 i \, e^{\left (8 i \, d x + 8 i \, c + \frac {7}{2}\right )} + 430 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 162 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 49 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} \sqrt {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 )} - 480 i \, \sqrt {2} e^{\left (7 i \, d x + 7 i \, c + \frac {7}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{1680 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1680*(sqrt(1/2)*(7*I*e^(7/2) - 15*I*e^(10*I*d*x + 10*I*c + 7/2) - 185*I*e^(8*I*d*x + 8*I*c + 7/2) + 430*I*e^
(6*I*d*x + 6*I*c + 7/2) + 162*I*e^(4*I*d*x + 4*I*c + 7/2) + 49*I*e^(2*I*d*x + 2*I*c + 7/2))*sqrt(e^(2*I*d*x +
2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) - 480*I*sqrt(2)*e^(7*I*d*x + 7*I*c + 7/2)*weierstrassPInverse(-4, 0, e^(I
*d*x + I*c)))*e^(-7*I*d*x - 7*I*c)/(a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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