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

Optimal. Leaf size=152 \[ -\frac{2 b \left (11 a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{5 d e}+\frac{2 a \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{e \cos (c+d x)}}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}-\frac{6 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e} \]

[Out]

(-2*b*(11*a^2 + 4*b^2)*Sqrt[e*Cos[c + d*x]])/(5*d*e) + (2*a*(a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*
x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]]) - (6*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x]))/(5*d*e) - (2*b*Sqrt[e*C
os[c + d*x]]*(a + b*Sin[c + d*x])^2)/(5*d*e)

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Rubi [A]  time = 0.24042, antiderivative size = 152, 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, 2642, 2641} \[ -\frac{2 b \left (11 a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{5 d e}+\frac{2 a \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{e \cos (c+d x)}}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}-\frac{6 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(11*a^2 + 4*b^2)*Sqrt[e*Cos[c + d*x]])/(5*d*e) + (2*a*(a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*
x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]]) - (6*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x]))/(5*d*e) - (2*b*Sqrt[e*C
os[c + d*x]]*(a + b*Sin[c + d*x])^2)/(5*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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{(a+b \sin (c+d x))^3}{\sqrt{e \cos (c+d x)}} \, dx &=-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}+\frac{2}{5} \int \frac{(a+b \sin (c+d x)) \left (\frac{5 a^2}{2}+2 b^2+\frac{9}{2} a b \sin (c+d x)\right )}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{6 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}+\frac{4}{15} \int \frac{\frac{15}{4} a \left (a^2+2 b^2\right )+\frac{3}{4} b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (11 a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{5 d e}-\frac{6 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}+\left (a \left (a^2+2 b^2\right )\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (11 a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{5 d e}-\frac{6 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}+\frac{\left (a \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{\sqrt{e \cos (c+d x)}}\\ &=-\frac{2 b \left (11 a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{5 d e}+\frac{2 a \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{e \cos (c+d x)}}-\frac{6 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\\ \end{align*}

Mathematica [A]  time = 0.771315, size = 94, normalized size = 0.62 \[ \frac{b \cos (c+d x) \left (-30 a^2-10 a b \sin (c+d x)+b^2 \cos (2 (c+d x))-9 b^2\right )+10 a \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*a*(a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + b*Cos[c + d*x]*(-30*a^2 - 9*b^2 + b^2*Cos[2
*(c + d*x)] - 10*a*b*Sin[c + d*x]))/(5*d*Sqrt[e*Cos[c + d*x]])

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Maple [A]  time = 1.382, size = 279, normalized size = 1.8 \begin{align*} -{\frac{2}{5\,d} \left ( 8\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-20\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-12\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+5\,\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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{3}+10\,\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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a{b}^{2}-30\,{a}^{2}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+10\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{3}+15\,{a}^{2}b\sin \left ( 1/2\,dx+c/2 \right ) +4\,{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]

-2/5/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(8*b^3*sin(1/2*d*x+1/2*c)^7-20*a*b^2*cos(1/2*d*x+1
/2*c)*sin(1/2*d*x+1/2*c)^4-12*b^3*sin(1/2*d*x+1/2*c)^5+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-
1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+10*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-30*a^2*b*sin(1/2*d*x+1/2*c)^3+10*a*b^2*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^2-4*sin(1/2*d*x+1/2*c)^3*b^3+15*a^2*b*sin(1/2*d*x+1/2*c)+4*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 \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\sqrt{e \cos \left (d x + c\right )}}\,{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((b*sin(d*x + c) + a)^3/sqrt(e*cos(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\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 )}}{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))/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\sqrt{e \cos \left (d x + c\right )}}\,{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="giac")

[Out]

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