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

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

[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 = 0.25, 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 = {2692, 2862, 2669, 2640, 2639} \[ \frac {2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac {2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}+\frac {2 b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))^2}{7 d e}+\frac {22 a b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))}{35 d e} \]

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 2692

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 2862

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] && Gt
Q[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 = 0.55, size = 105, normalized size = 0.65 \[ \frac {\sqrt {e \sin (c+d x)} \left (b \sin ^{\frac {3}{2}}(c+d x) \left (210 a^2+126 a b \cos (c+d x)+15 b^2 \cos (2 (c+d x))+55 b^2\right )-42 \left (5 a^3+6 a b^2\right ) E\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{105 d \sqrt {\sin (c+d x)}} \]

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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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}\right )} \sqrt {e \sin \left (d x + c\right )}, x\right ) \]

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]

integral((b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3)*sqrt(e*sin(d*x + c)), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.40, size = 315, normalized size = 1.96 \[ \frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (3 \left (\cos ^{2}\left (d x +c \right )\right ) b^{2}+21 a^{2}+4 b^{2}\right )}{21 e}-\frac {a e \left (10 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+12 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}+6 \left (\sin ^{4}\left (d x +c \right )\right ) b^{2}-6 \left (\sin ^{2}\left (d x +c \right )\right ) b^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

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)

[Out]

(2/21/e*b*(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 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {e \sin \left (d x + c\right )}\,{d x} \]

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]

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

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

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

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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