3.54 \(\int \frac{(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{2 b \left (3 a^2+4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e^3}-\frac{2 a \left (a^2+6 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))}{d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}} \]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*(a^2 + 6*b^2)*EllipticE[(c
- Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[c + d*x]]) - (2*b*(3*a^2 + 4*b^2)*(e*Sin[c + d*x])^(
3/2))/(3*d*e^3) - (2*a*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(d*e^3)

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Rubi [A]  time = 0.252763, antiderivative size = 165, 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 = {2691, 2862, 2669, 2640, 2639} \[ -\frac{2 b \left (3 a^2+4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e^3}-\frac{2 a \left (a^2+6 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 a b (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))}{d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*(a^2 + 6*b^2)*EllipticE[(c
- Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[c + d*x]]) - (2*b*(3*a^2 + 4*b^2)*(e*Sin[c + d*x])^(
3/2))/(3*d*e^3) - (2*a*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(d*e^3)

Rule 2691

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[((g*C
os[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(b + a*Sin[e + f*x]))/(f*g*(p + 1)), x] + Dist[1/(g^2*(p + 1
)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (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 \frac{(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{3/2}} \, dx &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}}-\frac{2 \int (a+b \cos (c+d x)) \left (\frac{a^2}{2}+2 b^2+\frac{5}{2} a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)} \, dx}{e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}}-\frac{2 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{d e^3}-\frac{4 \int \left (\frac{5}{4} a \left (a^2+6 b^2\right )+\frac{5}{4} b \left (3 a^2+4 b^2\right ) \cos (c+d x)\right ) \sqrt{e \sin (c+d x)} \, dx}{5 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}}-\frac{2 b \left (3 a^2+4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e^3}-\frac{2 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{d e^3}-\frac{\left (a \left (a^2+6 b^2\right )\right ) \int \sqrt{e \sin (c+d x)} \, dx}{e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}}-\frac{2 b \left (3 a^2+4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e^3}-\frac{2 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{d e^3}-\frac{\left (a \left (a^2+6 b^2\right ) \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{e^2 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{d e \sqrt{e \sin (c+d x)}}-\frac{2 a \left (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)}}{d e^2 \sqrt{\sin (c+d x)}}-\frac{2 b \left (3 a^2+4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e^3}-\frac{2 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{d e^3}\\ \end{align*}

Mathematica [A]  time = 0.322221, size = 101, normalized size = 0.61 \[ -\frac{2 \left (3 a \left (a^2+3 b^2\right ) \cos (c+d x)-3 a \left (a^2+6 b^2\right ) \sqrt{\sin (c+d x)} E\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+9 a^2 b+b^3 \sin ^2(c+d x)+3 b^3\right )}{3 d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(9*a^2*b + 3*b^3 + 3*a*(a^2 + 3*b^2)*Cos[c + d*x] - 3*a*(a^2 + 6*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2]*
Sqrt[Sin[c + d*x]] + b^3*Sin[c + d*x]^2))/(3*d*e*Sqrt[e*Sin[c + d*x]])

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Maple [A]  time = 2.81, size = 313, normalized size = 1.9 \begin{align*}{\frac{1}{3\,ed\cos \left ( dx+c \right ) } \left ( 6\,\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}^{3}+36\,\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{b}^{2}-3\,{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{a}^{3}-18\,{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }a{b}^{2}+2\,{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-6\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-18\,a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-18\,{a}^{2}b\cos \left ( dx+c \right ) -8\,{b}^{3}\cos \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \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))^(3/2),x)

[Out]

1/3/e/(e*sin(d*x+c))^(1/2)/cos(d*x+c)*(6*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Elliptic
E((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^3+36*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Ellipt
icE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a*b^2-3*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*(1-sin(d*x+c))^(1/2)
*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*a^3-18*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*(1-sin(d*x+c))^(1/
2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*a*b^2+2*b^3*cos(d*x+c)^3-6*a^3*cos(d*x+c)^2-18*a*b^2*cos(d*x+c)^2-1
8*a^2*b*cos(d*x+c)-8*b^3*cos(d*x+c))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\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 )}}{e^{2} \cos \left (d x + c\right )^{2} - e^{2}}, 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))^(3/2),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))/(e^2
*cos(d*x + c)^2 - e^2), 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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="giac")

[Out]

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