3.1.52 \(\int (a+b \cos (c+d x))^3 \sqrt {e \sin (c+d x)} \, dx\) [52]

Optimal. Leaf size=161 \[ \frac {2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}+\frac {2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac {22 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2}}{7 d e} \]

[Out]

2/105*b*(57*a^2+20*b^2)*(e*sin(d*x+c))^(3/2)/d/e+22/35*a*b*(a+b*cos(d*x+c))*(e*sin(d*x+c))^(3/2)/d/e+2/7*b*(a+
b*cos(d*x+c))^2*(e*sin(d*x+c))^(3/2)/d/e-2/5*a*(5*a^2+6*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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/sin(d*x+c)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 161, 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, 2719} \begin {gather*} \frac {2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac {2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}+\frac {2 b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))^2}{7 d e}+\frac {22 a b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))}{35 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*(5*a^2 + 6*b^2)*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) + (2*b*(5
7*a^2 + 20*b^2)*(e*Sin[c + d*x])^(3/2))/(105*d*e) + (22*a*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(35*d
*e) + (2*b*(a + b*Cos[c + d*x])^2*(e*Sin[c + d*x])^(3/2))/(7*d*e)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 (a+b \cos (c+d x))^3 \sqrt {e \sin (c+d x)} \, dx &=\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2}}{7 d e}+\frac {2}{7} \int (a+b \cos (c+d x)) \left (\frac {7 a^2}{2}+2 b^2+\frac {11}{2} a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)} \, dx\\ &=\frac {22 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2}}{7 d e}+\frac {4}{35} \int \left (\frac {7}{4} a \left (5 a^2+6 b^2\right )+\frac {1}{4} b \left (57 a^2+20 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)} \, dx\\ &=\frac {2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac {22 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2}}{7 d e}+\frac {1}{5} \left (a \left (5 a^2+6 b^2\right )\right ) \int \sqrt {e \sin (c+d x)} \, dx\\ &=\frac {2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac {22 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2}}{7 d e}+\frac {\left (a \left (5 a^2+6 b^2\right ) \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 \sqrt {\sin (c+d x)}}\\ &=\frac {2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}+\frac {2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac {22 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2}}{7 d e}\\ \end {align*}

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Mathematica [A]
time = 0.59, size = 105, normalized size = 0.65 \begin {gather*} \frac {\sqrt {e \sin (c+d x)} \left (-42 \left (5 a^3+6 a b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+b \left (210 a^2+55 b^2+126 a b \cos (c+d x)+15 b^2 \cos (2 (c+d x))\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{105 d \sqrt {\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]

(Sqrt[e*Sin[c + d*x]]*(-42*(5*a^3 + 6*a*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + b*(210*a^2 + 55*b^2 + 126*a
*b*Cos[c + d*x] + 15*b^2*Cos[2*(c + d*x)])*Sin[c + d*x]^(3/2)))/(105*d*Sqrt[Sin[c + d*x]])

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Maple [A]
time = 0.16, size = 315, normalized size = 1.96

method	result	size
default	\(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (3 \left (\cos ^{2}\left (d x +c \right )\right ) b^{2}+21 a^{2}+4 b^{2}\right )}{21 e}-\frac {a e \left (10 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+12 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}+6 \left (\sin ^{4}\left (d x +c \right )\right ) b^{2}-6 \left (\sin ^{2}\left (d x +c \right )\right ) b^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(315\)

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]

(2/21*b/e*(e*sin(d*x+c))^(3/2)*(3*cos(d*x+c)^2*b^2+21*a^2+4*b^2)-1/5*a*e*(10*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+
c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+12*(-sin(d*x+c)+1)^(1/2)*(2*sin(
d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-5*(-sin(d*x+c)+1)^(1/2)*(2*s
in(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2-6*(-sin(d*x+c)+1)^(1/2)*(
2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2+6*sin(d*x+c)^4*b^2-6*s
in(d*x+c)^2*b^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/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 = 0.12, size = 148, normalized size = 0.92 \begin {gather*} \frac {21 i \, \sqrt {2} \sqrt {-i} {\left (5 \, a^{3} + 6 \, a b^{2}\right )} e^{\frac {1}{2}} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} \sqrt {i} {\left (5 \, a^{3} + 6 \, a b^{2}\right )} e^{\frac {1}{2}} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (15 \, b^{3} \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} + 63 \, a b^{2} \cos \left (d x + c\right ) e^{\frac {1}{2}} + 5 \, {\left (21 \, a^{2} b + 4 \, b^{3}\right )} e^{\frac {1}{2}}\right )} \sin \left (d x + c\right )^{\frac {3}{2}}}{105 \, 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]

1/105*(21*I*sqrt(2)*sqrt(-I)*(5*a^3 + 6*a*b^2)*e^(1/2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x
+ c) + I*sin(d*x + c))) - 21*I*sqrt(2)*sqrt(I)*(5*a^3 + 6*a*b^2)*e^(1/2)*weierstrassZeta(4, 0, weierstrassPIn
verse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(15*b^3*cos(d*x + c)^2*e^(1/2) + 63*a*b^2*cos(d*x + c)*e^(1/2)
+ 5*(21*a^2*b + 4*b^3)*e^(1/2))*sin(d*x + c)^(3/2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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