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*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)

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Rubi [A]  time = 0.240711, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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])

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 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 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]

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*}

Mathematica [A]  time = 0.605945, 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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Maple [B]  time = 1.639, size = 339, normalized size = 2.2 \begin{align*}{\frac{2\,e}{105\,d} \left ( 240\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}-504\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-480\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-420\,{a}^{2}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+504\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+220\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+105\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{3}+126\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a{b}^{2}+420\,{a}^{2}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-126\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{3}-105\,{a}^{2}b\sin \left ( 1/2\,dx+c/2 \right ) -20\,{b}^{3}\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

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*sin(1/2*d*x+1/2*c)^3*b^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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \cos \left (d x + c\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out