3.1178 \(\int \frac{\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=261 \[ -\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-24 a b \sin (c+d x)-5 b^2\right )}{35 b^4 d}+\frac{8 \left (-37 a^2 b^2+32 a^4+5 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^5 d \sqrt{a+b \sin (c+d x)}}-\frac{8 a \left (32 a^2-29 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}} \]

[Out]

(-8*a*(32*a^2 - 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(35*b^5*d*Sqrt[
(a + b*Sin[c + d*x])/(a + b)]) + (8*(32*a^4 - 37*a^2*b^2 + 5*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]
*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(35*b^5*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*(8*a + b*Sin[c +
d*x]))/(7*b^2*d*Sqrt[a + b*Sin[c + d*x]]) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(32*a^2 - 5*b^2 - 24*a*b*
Sin[c + d*x]))/(35*b^4*d)

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Rubi [A]  time = 0.424217, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2863, 2865, 2752, 2663, 2661, 2655, 2653} \[ -\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-24 a b \sin (c+d x)-5 b^2\right )}{35 b^4 d}+\frac{8 \left (-37 a^2 b^2+32 a^4+5 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^5 d \sqrt{a+b \sin (c+d x)}}-\frac{8 a \left (32 a^2-29 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-8*a*(32*a^2 - 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(35*b^5*d*Sqrt[
(a + b*Sin[c + d*x])/(a + b)]) + (8*(32*a^4 - 37*a^2*b^2 + 5*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]
*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(35*b^5*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*(8*a + b*Sin[c +
d*x]))/(7*b^2*d*Sqrt[a + b*Sin[c + d*x]]) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(32*a^2 - 5*b^2 - 24*a*b*
Sin[c + d*x]))/(35*b^4*d)

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 2865

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 + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[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))^{3/2}} \, dx &=\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{12 \int \frac{\cos ^2(c+d x) \left (-\frac{b}{2}-4 a \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{7 b^2}\\ &=\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-5 b^2-24 a b \sin (c+d x)\right )}{35 b^4 d}-\frac{16 \int \frac{\frac{1}{4} b \left (8 a^2-5 b^2\right )+\frac{1}{4} a \left (32 a^2-29 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{35 b^4}\\ &=\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-5 b^2-24 a b \sin (c+d x)\right )}{35 b^4 d}-\frac{\left (4 a \left (32 a^2-29 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{35 b^5}+\frac{\left (4 \left (32 a^4-37 a^2 b^2+5 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{35 b^5}\\ &=\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-5 b^2-24 a b \sin (c+d x)\right )}{35 b^4 d}-\frac{\left (4 a \left (32 a^2-29 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}{35 b^5 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (4 \left (32 a^4-37 a^2 b^2+5 b^4\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}{35 b^5 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{8 a \left (32 a^2-29 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)}}{35 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 \left (32 a^4-37 a^2 b^2+5 b^4\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}}}{35 b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \cos ^3(c+d x) (8 a+b \sin (c+d x))}{7 b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-5 b^2-24 a b \sin (c+d x)\right )}{35 b^4 d}\\ \end{align*}

Mathematica [A]  time = 3.91684, size = 222, normalized size = 0.85 \[ \frac{b \cos (c+d x) \left (\left (45 b^3-64 a^2 b\right ) \sin (c+d x)-256 a^3-16 a b^2 \cos (2 (c+d x))+216 a b^2+5 b^3 \sin (3 (c+d x))\right )-16 \left (-37 a^2 b^2+32 a^4+5 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+16 a \left (32 a^2 b+32 a^3-29 a b^2-29 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{70 b^5 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(16*a*(32*a^3 + 32*a^2*b - 29*a*b^2 - 29*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[
c + d*x])/(a + b)] - 16*(32*a^4 - 37*a^2*b^2 + 5*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)] + b*Cos[c + d*x]*(-256*a^3 + 216*a*b^2 - 16*a*b^2*Cos[2*(c + d*x)] + (-64*a^2*b + 4
5*b^3)*Sin[c + d*x] + 5*b^3*Sin[3*(c + d*x)]))/(70*b^5*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 1.546, size = 943, normalized size = 3.6 \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))^(3/2),x)

[Out]

-2/35*(-5*b^5*sin(d*x+c)^5+128*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))
*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b-96*((a+b*sin(d*x+c))/(a-b)
)^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/
2),((a-b)/(a+b))^(1/2))*a^3*b^2-148*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*
x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3+96*((a+b*sin(d*x+c)
)/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-
b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4+20*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+si
n(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^5-128*((a+b*sin(d*x+c
))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a
-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5+244*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+si
n(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-116*((a+b*sin(d
*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c)
)/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4+8*a*b^4*sin(d*x+c)^4-16*a^2*b^3*sin(d*x+c)^3+20*b^5*sin(d*x+c)^3-64*
a^3*b^2*sin(d*x+c)^2+42*a*b^4*sin(d*x+c)^2+16*a^2*b^3*sin(d*x+c)-15*b^5*sin(d*x+c)+64*a^3*b^2-50*a*b^4)/b^6/co
s(d*x+c)/(a+b*sin(d*x+c))^(1/2)/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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)/(b*sin(d*x + c) + a)^(3/2), 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 )}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, 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))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^4*sin(d*x + c)/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2
- b^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(cos(d*x+c)**4*sin(d*x+c)/(a+b*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{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)/(b*sin(d*x + c) + a)^(3/2), x)