3.561 \(\int \frac{(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=164 \[ \frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 a \left (a^2-6 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt{e \cos (c+d x)}}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d e (e \cos (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.249331, antiderivative size = 164, 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, 2642, 2641} \[ \frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 a \left (a^2-6 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt{e \cos (c+d x)}}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.684329, size = 103, normalized size = 0.63 \[ \frac{2 a \left (a^2-6 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 a^2 b+2 a^3 \sin (c+d x)+6 a b^2 \sin (c+d x)+3 b^3 \cos (2 (c+d x))+5 b^3}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(5/2),x]

[Out]

(6*a^2*b + 5*b^3 + 3*b^3*Cos[2*(c + d*x)] + 2*a*(a^2 - 6*b^2)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 2
*a^3*Sin[c + d*x] + 6*a*b^2*Sin[c + d*x])/(3*d*e*(e*Cos[c + d*x])^(3/2))

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Maple [B]  time = 1.881, size = 384, normalized size = 2.3 \begin{align*} -{\frac{2}{3\,d{e}^{2}} \left ( 2\,\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} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-12\,\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} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+12\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ){a}^{3}+6\,\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}+2\,{a}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+6\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-12\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{3}+3\,{a}^{2}b\sin \left ( 1/2\,dx+c/2 \right ) +4\,{b}^{3}\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \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))^(5/2),x)

[Out]

-2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^2*(2*(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*sin(1/2*d*x+1/2*c)^2-1
2*(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*si
n(1/2*d*x+1/2*c)^2+12*b^3*sin(1/2*d*x+1/2*c)^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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+6*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+2*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+6*a*b^2*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^2-12*sin(1/2*d*x+1/2*c)^3*b^3+3*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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(5/2), 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^{3} \cos \left (d x + c\right )^{3}}, 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))^(5/2),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^3*cos(d*x + c)^3), 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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(5/2), x)