3.1185 \(\int \frac{\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=254 \[ \frac{4 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{15 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{8 a \left (32 a^2-17 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{8 \left (32 a^2-9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}} \]

[Out]

(8*(32*a^2 - 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*b^5*d*Sqrt[(a +
b*Sin[c + d*x])/(a + b)]) - (8*a*(32*a^2 - 17*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)])/(15*b^5*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*(8*a + 3*b*Sin[c + d*x]))/(15*b
^2*d*(a + b*Sin[c + d*x])^(3/2)) + (4*Cos[c + d*x]*(32*a^2 - 9*b^2 + 8*a*b*Sin[c + d*x]))/(15*b^4*d*Sqrt[a + b
*Sin[c + d*x]])

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Rubi [A]  time = 0.440458, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2863, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 \cos (c+d x) \left (32 a^2+8 a b \sin (c+d x)-9 b^2\right )}{15 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{8 a \left (32 a^2-17 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{8 \left (32 a^2-9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(8*(32*a^2 - 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*b^5*d*Sqrt[(a +
b*Sin[c + d*x])/(a + b)]) - (8*a*(32*a^2 - 17*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)])/(15*b^5*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*(8*a + 3*b*Sin[c + d*x]))/(15*b
^2*d*(a + b*Sin[c + d*x])^(3/2)) + (4*Cos[c + d*x]*(32*a^2 - 9*b^2 + 8*a*b*Sin[c + d*x]))/(15*b^4*d*Sqrt[a + b
*Sin[c + d*x]])

Rule 2863

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}-\frac{4 \int \frac{\cos ^2(c+d x) \left (-\frac{3 b}{2}-4 a \sin (c+d x)\right )}{(a+b \sin (c+d x))^{3/2}} \, dx}{5 b^2}\\ &=\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x) \left (32 a^2-9 b^2+8 a b \sin (c+d x)\right )}{15 b^4 d \sqrt{a+b \sin (c+d x)}}+\frac{16 \int \frac{2 a b+\frac{1}{4} \left (32 a^2-9 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 b^4}\\ &=\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x) \left (32 a^2-9 b^2+8 a b \sin (c+d x)\right )}{15 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{\left (4 a \left (32 a^2-17 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 b^5}+\frac{\left (4 \left (32 a^2-9 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{15 b^5}\\ &=\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x) \left (32 a^2-9 b^2+8 a b \sin (c+d x)\right )}{15 b^4 d \sqrt{a+b \sin (c+d x)}}+\frac{\left (4 \left (32 a^2-9 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{15 b^5 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (4 a \left (32 a^2-17 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{15 b^5 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{8 \left (32 a^2-9 b^2\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{15 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{8 a \left (32 a^2-17 b^2\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{15 b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \cos ^3(c+d x) (8 a+3 b \sin (c+d x))}{15 b^2 d (a+b \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x) \left (32 a^2-9 b^2+8 a b \sin (c+d x)\right )}{15 b^4 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.8889, size = 211, normalized size = 0.83 \[ \frac{2 b \cos (c+d x) \left (b \left (320 a^2-69 b^2\right ) \sin (c+d x)+256 a^3-16 a b^2 \cos (2 (c+d x))-24 a b^2+3 b^3 \sin (3 (c+d x))\right )+32 a \left (32 a^2-17 b^2\right ) (a+b) \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-32 \left (32 a^2-9 b^2\right ) (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{60 b^5 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-32*(a + b)^2*(32*a^2 - 9*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))
^(3/2) + 32*a*(a + b)*(32*a^2 - 17*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/
(a + b))^(3/2) + 2*b*Cos[c + d*x]*(256*a^3 - 24*a*b^2 - 16*a*b^2*Cos[2*(c + d*x)] + b*(320*a^2 - 69*b^2)*Sin[c
+ d*x] + 3*b^3*Sin[3*(c + d*x)]))/(60*b^5*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [B]  time = 1.565, size = 1430, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x)

[Out]

2/15*(3*b^5*sin(d*x+c)*cos(d*x+c)^4+(80*a^2*b^3-18*b^5)*cos(d*x+c)^2*sin(d*x+c)+4*(-b/(a-b)*sin(d*x+c)-b/(a-b)
)^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*b*(32*EllipticF((b/(a-b)*sin(
d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^3*b-24*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a
+b))^(1/2))*a^2*b^2-17*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a*b^3+9*EllipticF((
b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*b^4-32*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)
,((a-b)/(a+b))^(1/2))*a^4+41*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^2-9*Ell
ipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*b^4)*sin(d*x+c)-8*a*b^4*cos(d*x+c)^4+(64*a^3*
b^2-2*a*b^4)*cos(d*x+c)^2+128*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/
2),((a-b)/(a+b))^(1/2))*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^4*b-96*(-b/
(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(b/(a-b)*s
in(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^3*b^2-68*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*
EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(
a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^2*b^3+36*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/
(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a
*b^4-128*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+c
)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^5+164*(-b/(a-b)*sin(d*x+c)-b/(a-
b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1
/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^3*b^2-36*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/
(a-b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b
))^(1/2)*a*b^4)/(a+b*sin(d*x+c))^(3/2)/b^6/cos(d*x+c)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)/(b*sin(d*x + c) + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{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 )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^4*sin(d*x + c)/(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)), 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(cos(d*x+c)**4*sin(d*x+c)/(a+b*sin(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{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)/(b*sin(d*x + c) + a)^(5/2), x)