3.3.61 \(\int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3} \, dx\) [261]

Optimal. Leaf size=153 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 a^3 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac {10 \sqrt {e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac {10 \sqrt {e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

10/77*(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)/a
^3/d/(e*cos(d*x+c))^(1/2)-2/11*(e*cos(d*x+c))^(1/2)/d/e/(a+a*sin(d*x+c))^3-10/77*(e*cos(d*x+c))^(1/2)/a/d/e/(a
+a*sin(d*x+c))^2-10/77*(e*cos(d*x+c))^(1/2)/d/e/(a^3+a^3*sin(d*x+c))

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Rubi [A]
time = 0.13, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2760, 2762, 2721, 2720} \begin {gather*} -\frac {10 \sqrt {e \cos (c+d x)}}{77 d e \left (a^3 \sin (c+d x)+a^3\right )}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 a^3 d \sqrt {e \cos (c+d x)}}-\frac {10 \sqrt {e \cos (c+d x)}}{77 a d e (a \sin (c+d x)+a)^2}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^3),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*a^3*d*Sqrt[e*Cos[c + d*x]]) - (2*Sqrt[e*Cos[c + d*x]])/(
11*d*e*(a + a*Sin[c + d*x])^3) - (10*Sqrt[e*Cos[c + d*x]])/(77*a*d*e*(a + a*Sin[c + d*x])^2) - (10*Sqrt[e*Cos[
c + d*x]])/(77*d*e*(a^3 + a^3*Sin[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 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 2760

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

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}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3} \, dx &=-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}+\frac {5 \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx}{11 a}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac {10 \sqrt {e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}+\frac {15 \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))} \, dx}{77 a^2}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac {10 \sqrt {e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac {10 \sqrt {e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )}+\frac {5 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{77 a^3}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac {10 \sqrt {e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac {10 \sqrt {e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )}+\frac {\left (5 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 a^3 \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 )}{77 a^3 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac {10 \sqrt {e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac {10 \sqrt {e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.05, size = 66, normalized size = 0.43 \begin {gather*} -\frac {\sqrt {e \cos (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {15}{4};\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{2\ 2^{3/4} a^3 d e \sqrt [4]{1+\sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^3),x]

[Out]

-1/2*(Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[1/4, 15/4, 5/4, (1 - Sin[c + d*x])/2])/(2^(3/4)*a^3*d*e*(1 + Sin[
c + d*x])^(1/4))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(579\) vs. \(2(161)=322\).
time = 10.26, size = 580, normalized size = 3.79

method	result	size
default	\(-\frac {2 \left (160 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-400 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+400 \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 )-320 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-200 \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 )+264 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+50 \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 )-104 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+44 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )+72 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-44 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-17 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{77 \left (32 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-80 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{3} \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}\)	\(580\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/77/(32*sin(1/2*d*x+1/2*c)^10-80*sin(1/2*d*x+1/2*c)^8+80*sin(1/2*d*x+1/2*c)^6-40*sin(1/2*d*x+1/2*c)^4+10*sin
(1/2*d*x+1/2*c)^2-1)/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(160*EllipticF(cos(1/2*d*x+1/2
*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^10-400*EllipticF
(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^
8+160*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+400*(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)^6-320*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-200*
(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+264*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+50*(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)^2-104*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*
c)+44*sin(1/2*d*x+1/2*c)^5-5*(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))+72*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-44*sin(1/2*d*x+1/2*c)^3-17*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/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

e^(-1/2)*integrate(1/((a*sin(d*x + c) + a)^3*sqrt(cos(d*x + c))), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 221, normalized size = 1.44 \begin {gather*} -\frac {5 \, {\left (3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 4 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) - 4 i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 4 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) + 4 i \, \sqrt {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} - 15 \, \sin \left (d x + c\right ) - 22\right )} \sqrt {\cos \left (d x + c\right )}}{77 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} - 4 \, a^{3} d e^{\frac {1}{2}} + {\left (a^{3} d \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} - 4 \, a^{3} d e^{\frac {1}{2}}\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/77*(5*(3*I*sqrt(2)*cos(d*x + c)^2 + (I*sqrt(2)*cos(d*x + c)^2 - 4*I*sqrt(2))*sin(d*x + c) - 4*I*sqrt(2))*we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(-3*I*sqrt(2)*cos(d*x + c)^2 + (-I*sqrt(2)*cos(d*x
+ c)^2 + 4*I*sqrt(2))*sin(d*x + c) + 4*I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) +
2*(5*cos(d*x + c)^2 - 15*sin(d*x + c) - 22)*sqrt(cos(d*x + c)))/(3*a^3*d*cos(d*x + c)^2*e^(1/2) - 4*a^3*d*e^(
1/2) + (a^3*d*cos(d*x + c)^2*e^(1/2) - 4*a^3*d*e^(1/2))*sin(d*x + c))

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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/(a+a*sin(d*x+c))**3/(e*cos(d*x+c))**(1/2),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(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(e^(-1/2)/((a*sin(d*x + c) + a)^3*sqrt(cos(d*x + c))), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {e\,\cos \left (c+d\,x\right )}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cos(c + d*x))^(1/2)*(a + a*sin(c + d*x))^3),x)

[Out]

int(1/((e*cos(c + d*x))^(1/2)*(a + a*sin(c + d*x))^3), x)

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