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

Optimal. Leaf size=160 \[ \frac{2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac{2 a \left (a^2+6 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{d e^2 \sqrt{\cos (c+d x)}}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt{e \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.241024, antiderivative size = 160, 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} \[ \frac{2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac{2 a \left (a^2+6 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{d e^2 \sqrt{\cos (c+d x)}}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.396142, size = 98, normalized size = 0.61 \[ \frac{6 \left (a \left (a^2+3 b^2\right ) \sin (c+d x)+3 a^2 b+b^3\right )-6 a \left (a^2+6 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 b^3 \cos ^2(c+d x)}{3 d e \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.937, size = 248, normalized size = 1.6 \begin{align*} -{\frac{2}{3\,de} \left ( -4\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+3\,\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}+18\,\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}-6\,{a}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-18\,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}-9\,{a}^{2}b\sin \left ( 1/2\,dx+c/2 \right ) -4\,{b}^{3}\sin \left ( 1/2\,dx+c/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \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))^(3/2),x)

[Out]

-2/3/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)*(-4*b^3*sin(1/2*d*x+1/2*c)^5+3*(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+18*(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^2-6*a^3*cos(1/2*d*x+1/2*c)
*sin(1/2*d*x+1/2*c)^2-18*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+4*sin(1/2*d*x+1/2*c)^3*b^3-9*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}}{\left (e \cos \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*sin(d*x+c))^3/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^3/(e*cos(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 (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^{2} \cos \left (d x + c\right )^{2}}, 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))^(3/2),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^2*cos(d*x + c)^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*sin(d*x+c))**3/(e*cos(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 \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \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*sin(d*x+c))^3/(e*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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