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

Optimal. Leaf size=140 \[ -\frac {22 a^3 (e \cos (c+d x))^{3/2}}{15 d e}-\frac {22 \left (a^3 \sin (c+d x)+a^3\right ) (e \cos (c+d x))^{3/2}}{35 d e}+\frac {22 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}{7 d e} \]

[Out]

-22/15*a^3*(e*cos(d*x+c))^(3/2)/d/e-2/7*a*(e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^2/d/e-22/35*(e*cos(d*x+c))^(3/
2)*(a^3+a^3*sin(d*x+c))/d/e+22/5*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2
*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 140, 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 = {2678, 2669, 2640, 2639} \[ -\frac {22 a^3 (e \cos (c+d x))^{3/2}}{15 d e}-\frac {22 \left (a^3 \sin (c+d x)+a^3\right ) (e \cos (c+d x))^{3/2}}{35 d e}+\frac {22 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}{7 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-22*a^3*(e*Cos[c + d*x])^(3/2))/(15*d*e) + (22*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[
Cos[c + d*x]]) - (2*a*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2)/(7*d*e) - (22*(e*Cos[c + d*x])^(3/2)*(a^3
+ a^3*Sin[c + d*x]))/(35*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 2678

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[(a*(2*m + p - 1))/(m + p), 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]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3 \, dx &=-\frac {2 a (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}{7 d e}+\frac {1}{7} (11 a) \int \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {2 a (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}{7 d e}-\frac {22 (e \cos (c+d x))^{3/2} \left (a^3+a^3 \sin (c+d x)\right )}{35 d e}+\frac {1}{5} \left (11 a^2\right ) \int \sqrt {e \cos (c+d x)} (a+a \sin (c+d x)) \, dx\\ &=-\frac {22 a^3 (e \cos (c+d x))^{3/2}}{15 d e}-\frac {2 a (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}{7 d e}-\frac {22 (e \cos (c+d x))^{3/2} \left (a^3+a^3 \sin (c+d x)\right )}{35 d e}+\frac {1}{5} \left (11 a^3\right ) \int \sqrt {e \cos (c+d x)} \, dx\\ &=-\frac {22 a^3 (e \cos (c+d x))^{3/2}}{15 d e}-\frac {2 a (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}{7 d e}-\frac {22 (e \cos (c+d x))^{3/2} \left (a^3+a^3 \sin (c+d x)\right )}{35 d e}+\frac {\left (11 a^3 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=-\frac {22 a^3 (e \cos (c+d x))^{3/2}}{15 d e}+\frac {22 a^3 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}-\frac {2 a (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}{7 d e}-\frac {22 (e \cos (c+d x))^{3/2} \left (a^3+a^3 \sin (c+d x)\right )}{35 d e}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 66, normalized size = 0.47 \[ -\frac {16\ 2^{3/4} a^3 (e \cos (c+d x))^{3/2} \, _2F_1\left (-\frac {11}{4},\frac {3}{4};\frac {7}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{3 d e (\sin (c+d x)+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*2^(3/4)*a^3*(e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[-11/4, 3/4, 7/4, (1 - Sin[c + d*x])/2])/(3*d*e*(1 +
Sin[c + d*x])^(3/4))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*sqrt(e*cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.85, size = 214, normalized size = 1.53 \[ \frac {2 a^{3} e \left (240 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-504 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-480 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-200 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+231 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-126 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+440 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-125 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*e*(240*sin(1/2*d*x+1/2*c)^9-504*sin(1/2*d*x+1
/2*c)^6*cos(1/2*d*x+1/2*c)-480*sin(1/2*d*x+1/2*c)^7+504*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-200*sin(1/2*d*
x+1/2*c)^5+231*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(
1/2)-126*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+440*sin(1/2*d*x+1/2*c)^3-125*sin(1/2*d*x+1/2*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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