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*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)

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Rubi [A]  time = 0.24989, antiderivative size = 161, 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} \[ \frac{2 b \left (57 a^2+20 b^2\right ) (e \sin (c+d x))^{3/2}}{105 d e}+\frac{2 a \left (5 a^2+6 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d \sqrt{\sin (c+d x)}}+\frac{2 b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))^2}{7 d e}+\frac{22 a b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))}{35 d e} \]

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 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 (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*}

Mathematica [A]  time = 0.553626, 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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Maple [A]  time = 2.518, size = 315, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({\frac{2\,b \left ( 3\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,{a}^{2}+4\,{b}^{2} \right ) }{21\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{ea}{5\,\cos \left ( dx+c \right ) } \left ( 10\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}+12\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{2}-5\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}-6\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{2}+6\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}{b}^{2}-6\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{2} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \end{align*}

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*b^2*cos(d*x+c)^2+21*a^2+4*b^2)-1/5*a*e*(10*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x
+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+12*(1-sin(d*x+c))^(1/2)*(2+2*sin(d
*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-5*(1-sin(d*x+c))^(1/2)*(2+2*sin(
d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-6*(1-sin(d*x+c))^(1/2)*(2+2*sin
(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+6*sin(d*x+c)^4*b^2-6*sin(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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt{e \sin \left (d x + c\right )}\,{d x} \end{align*}

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

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

[Out]

Timed out