3.3.41 \(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx\) [241]

Optimal. Leaf size=112 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \]

[Out]

10/21*sin(d*x+c)/a/d/e/(e*cos(d*x+c))^(3/2)-2/7/d/e/(e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))+10/21*(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/d/e^2/(e*cos(d*x+c)
)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2762, 2716, 2721, 2720} \begin {gather*} \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*a*d*e^2*Sqrt[e*Cos[c + d*x]]) + (10*Sin[c + d*x])/(21*a*
d*e*(e*Cos[c + d*x])^(3/2)) - 2/(7*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x]))

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2762

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((g*Cos[e
+ f*x])^(p + 1)/(a*f*g*(p - 1)*(a + b*Sin[e + f*x]))), x] + Dist[p/(a*(p - 1)), Int[(g*Cos[e + f*x])^p, x], x
] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx &=-\frac {2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}+\frac {5 \int \frac {1}{(e \cos (c+d x))^{5/2}} \, dx}{7 a}\\ &=\frac {10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}+\frac {5 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{21 a e^2}\\ &=\frac {10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}+\frac {\left (5 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 a e^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.07, size = 66, normalized size = 0.59 \begin {gather*} \frac {\, _2F_1\left (-\frac {3}{4},\frac {11}{4};\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{3\ 2^{3/4} a d e (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])),x]

[Out]

(Hypergeometric2F1[-3/4, 11/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4))/(3*2^(3/4)*a*d*e*(e*Cos[c
+ d*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(374\) vs. \(2(124)=248\).
time = 6.18, size = 375, normalized size = 3.35

method	result	size
default	\(-\frac {2 \left (40 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-60 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(375\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/a/sin(1/2*d*x+1/2*c)/(-2*sin(1
/2*d*x+1/2*c)^2*e+e)^(1/2)/e^2*(40*(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))*sin(1/2*d*x+1/2*c)^6-60*(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))*sin(1/2*d*x+1/2*c)^4+40*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+30*(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))*sin(1/2*d*x+1/
2*c)^2-40*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-5*(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))+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*sin(1/2*d*x+1/2*c))/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate(1/((a*sin(d*x + c) + a)*cos(d*x + c)^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 174, normalized size = 1.55 \begin {gather*} -\frac {5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )}}{21 \, {\left (a d \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2} e^{\frac {5}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/21*(5*(I*sqrt(2)*cos(d*x + c)^2*sin(d*x + c) + I*sqrt(2)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c)) + 5*(-I*sqrt(2)*cos(d*x + c)^2*sin(d*x + c) - I*sqrt(2)*cos(d*x + c)^2)*weierstrassPIn
verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(5*cos(d*x + c)^2 - 5*sin(d*x + c) - 2)*sqrt(cos(d*x + c)))/(a
*d*cos(d*x + c)^2*e^(5/2)*sin(d*x + c) + a*d*cos(d*x + c)^2*e^(5/2))

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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(1/(e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c)),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(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate(e^(-5/2)/((a*sin(d*x + c) + a)*cos(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))),x)

[Out]

int(1/((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))), x)

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