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

Optimal. Leaf size=156 \[ -\frac {2 b \left (57 a^2+20 b^2\right ) (e \cos (c+d x))^{3/2}}{105 d e}+\frac {2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}-\frac {22 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{35 d e} \]

[Out]

-2/105*b*(57*a^2+20*b^2)*(e*cos(d*x+c))^(3/2)/d/e-22/35*a*b*(e*cos(d*x+c))^(3/2)*(a+b*sin(d*x+c))/d/e-2/7*b*(e
*cos(d*x+c))^(3/2)*(a+b*sin(d*x+c))^2/d/e+2/5*a*(5*a^2+6*b^2)*(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 = 0.24, antiderivative size = 156, 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} \[ -\frac {2 b \left (57 a^2+20 b^2\right ) (e \cos (c+d x))^{3/2}}{105 d e}+\frac {2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}-\frac {22 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{35 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(57*a^2 + 20*b^2)*(e*Cos[c + d*x])^(3/2))/(105*d*e) + (2*a*(5*a^2 + 6*b^2)*Sqrt[e*Cos[c + d*x]]*Elliptic
E[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) - (22*a*b*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x]))/(35*d*e) -
(2*b*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x])^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 \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3 \, dx &=-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}+\frac {2}{7} \int \sqrt {e \cos (c+d x)} (a+b \sin (c+d x)) \left (\frac {7 a^2}{2}+2 b^2+\frac {11}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac {22 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{35 d e}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}+\frac {4}{35} \int \sqrt {e \cos (c+d x)} \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 ) \sin (c+d x)\right ) \, dx\\ &=-\frac {2 b \left (57 a^2+20 b^2\right ) (e \cos (c+d x))^{3/2}}{105 d e}-\frac {22 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{35 d e}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}+\frac {1}{5} \left (a \left (5 a^2+6 b^2\right )\right ) \int \sqrt {e \cos (c+d x)} \, dx\\ &=-\frac {2 b \left (57 a^2+20 b^2\right ) (e \cos (c+d x))^{3/2}}{105 d e}-\frac {22 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{35 d e}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}+\frac {\left (a \left (5 a^2+6 b^2\right ) \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=-\frac {2 b \left (57 a^2+20 b^2\right ) (e \cos (c+d x))^{3/2}}{105 d e}+\frac {2 a \left (5 a^2+6 b^2\right ) \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 {22 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{35 d e}-\frac {2 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{7 d e}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 101, normalized size = 0.65 \[ \frac {\sqrt {e \cos (c+d x)} \left (42 \left (5 a^3+6 a b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \cos ^{\frac {3}{2}}(c+d x) \left (-210 a^2-126 a b \sin (c+d x)+15 b^2 \cos (2 (c+d x))-55 b^2\right )\right )}{105 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.12, size = 339, normalized size = 2.17 \[ \frac {2 e \left (240 b^{3} \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-504 a \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 b^{3} \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-420 a^{2} b \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 a \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+220 b^{3} \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}+126 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}+420 a^{2} b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-126 a \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 b^{3} \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-105 a^{2} b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-20 b^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*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)*e*(240*b^3*sin(1/2*d*x+1/2*c)^9-504*a*b^2*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-480*b^3*sin(1/2*d*x+1/2*c)^7-420*a^2*b*sin(1/2*d*x+1/2*c)^5+504*a*b^2*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+220*b^3*sin(1/2*d*x+1/2*c)^5+105*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+126*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+420*a^2*b*sin(1/2*d*x+1/2*c)^3-126*a*b^2*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+20*b^3*sin(1/2*d*x+1/2*c)^3-105*a^2*b*sin(1/2*d*x+1/2*c)-20*b^3*sin(1/2*d*x+1/2
*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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