3.1.53 \(\int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx\) [53]

Optimal. Leaf size=157 \[ \frac {2 a \left (a^2+2 b^2\right ) F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {6 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e} \]

[Out]

-2*a*(a^2+2*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*
d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)+2/5*b*(11*a^2+4*b^2)*(e*sin(d*x+c))^(1/2)/d/e+6/5*a*b*(a
+b*cos(d*x+c))*(e*sin(d*x+c))^(1/2)/d/e+2/5*b*(a+b*cos(d*x+c))^2*(e*sin(d*x+c))^(1/2)/d/e

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Rubi [A]
time = 0.16, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2771, 2941, 2748, 2721, 2720} \begin {gather*} \frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 a \left (a^2+2 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{5 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2 + 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x]]) + (2*b*(11*a^
2 + 4*b^2)*Sqrt[e*Sin[c + d*x]])/(5*d*e) + (6*a*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(5*d*e) + (2*b*(a
+ b*Cos[c + d*x])^2*Sqrt[e*Sin[c + d*x]])/(5*d*e)

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 2748

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

Rule 2771

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

Rule 2941

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

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx &=\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (\frac {5 a^2}{2}+2 b^2+\frac {9}{2} a b \cos (c+d x)\right )}{\sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {6 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e}+\frac {4}{15} \int \frac {\frac {15}{4} a \left (a^2+2 b^2\right )+\frac {3}{4} b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {6 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e}+\left (a \left (a^2+2 b^2\right )\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {6 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e}+\frac {\left (a \left (a^2+2 b^2\right ) \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{\sqrt {e \sin (c+d x)}}\\ &=\frac {2 a \left (a^2+2 b^2\right ) F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {6 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 98, normalized size = 0.62 \begin {gather*} \frac {-10 a \left (a^2+2 b^2\right ) F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sqrt {\sin (c+d x)}+b \left (30 a^2+9 b^2+10 a b \cos (c+d x)+b^2 \cos (2 (c+d x))\right ) \sin (c+d x)}{5 d \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*a*(a^2 + 2*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sqrt[Sin[c + d*x]] + b*(30*a^2 + 9*b^2 + 10*a*b*Cos[c
+ d*x] + b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(5*d*Sqrt[e*Sin[c + d*x]])

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Maple [A]
time = 0.13, size = 210, normalized size = 1.34

method	result	size
default	\(-\frac {5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{3}+10 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}-2 b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 a \,b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-30 a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(210\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(5*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Elliptic
F((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^3+10*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Elli
pticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a*b^2-2*b^3*cos(d*x+c)^3*sin(d*x+c)-10*a*b^2*cos(d*x+c)^2*sin(d*x+c)-
30*a^2*b*cos(d*x+c)*sin(d*x+c)-8*b^3*cos(d*x+c)*sin(d*x+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((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

e^(-1/2)*integrate((b*cos(d*x + c) + a)^3/sqrt(sin(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 = 126, normalized size = 0.80 \begin {gather*} \frac {{\left (5 \, \sqrt {2} \sqrt {-i} {\left (a^{3} + 2 \, a b^{2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} \sqrt {i} {\left (a^{3} + 2 \, a b^{2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} + 5 \, a b^{2} \cos \left (d x + c\right ) + 15 \, a^{2} b + 4 \, b^{3}\right )} \sqrt {\sin \left (d x + c\right )}\right )} e^{\left (-\frac {1}{2}\right )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*sqrt(2)*sqrt(-I)*(a^3 + 2*a*b^2)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*s
qrt(I)*(a^3 + 2*a*b^2)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(b^3*cos(d*x + c)^2 + 5*a*
b^2*cos(d*x + c) + 15*a^2*b + 4*b^3)*sqrt(sin(d*x + c)))*e^(-1/2)/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \cos {\left (c + d x \right )}\right )^{3}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*cos(c + d*x))**3/sqrt(e*sin(c + d*x)), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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