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

Optimal. Leaf size=92 \[ -\frac {2 \cos ^5(c+d x)}{9 a d \sqrt {a \sin (c+d x)+a}}+\frac {20 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac {46 a \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

-46/315*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)+20/63*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/9*cos(d*x+c)^5/a
/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2877, 2856, 2674, 2673} \[ -\frac {2 \cos ^5(c+d x)}{9 a d \sqrt {a \sin (c+d x)+a}}+\frac {20 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac {46 a \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-46*a*Cos[c + d*x]^5)/(315*d*(a + a*Sin[c + d*x])^(5/2)) + (20*Cos[c + d*x]^5)/(63*d*(a + a*Sin[c + d*x])^(3/
2)) - (2*Cos[c + d*x]^5)/(9*a*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2856

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[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

Rule 2877

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m + p + 1)), x] - Dist[1/(a^
2*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2*m + p + 1)*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2^(-1)] && NeQ[2*m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\cos ^5(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {\int \frac {\cos ^4(c+d x) \left (-\frac {3 a}{2}-2 a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac {\cos ^5(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 a d \sqrt {a+a \sin (c+d x)}}+\frac {23 \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{36 a}\\ &=\frac {20 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 a d \sqrt {a+a \sin (c+d x)}}+\frac {23}{63} \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {46 a \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac {20 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 a d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.72, size = 92, normalized size = 1.00 \[ -\frac {\sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 (40 \sin (c+d x)-35 \cos (2 (c+d x))+51)}{315 a^2 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/315*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(51 - 35*Cos[2*(c + d*x)] + 40*Sin[
c + d*x]))/(a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.44, size = 142, normalized size = 1.54 \[ -\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} + 85 \, \cos \left (d x + c\right )^{4} - 73 \, \cos \left (d x + c\right )^{3} - 169 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{3} - 123 \, \cos \left (d x + c\right )^{2} + 46 \, \cos \left (d x + c\right ) + 92\right )} \sin \left (d x + c\right ) + 46 \, \cos \left (d x + c\right ) + 92\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2/315*(35*cos(d*x + c)^5 + 85*cos(d*x + c)^4 - 73*cos(d*x + c)^3 - 169*cos(d*x + c)^2 - (35*cos(d*x + c)^4 -
50*cos(d*x + c)^3 - 123*cos(d*x + c)^2 + 46*cos(d*x + c) + 92)*sin(d*x + c) + 46*cos(d*x + c) + 92)*sqrt(a*sin
(d*x + c) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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giac [B]  time = 0.72, size = 282, normalized size = 3.07 \[ \frac {8 \, {\left (\frac {23 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}}} - \frac {{\left (\frac {9 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {105 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {252 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {252 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {105 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {9 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {2 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}}}\right )}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

8/315*(23*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(3/2) - ((9*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (105*a^3/sgn
(tan(1/2*d*x + 1/2*c) + 1) - (252*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (252*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) -
(105*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (2*a^3*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c) + 1) + 9*a^3/
sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x +
1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^2 + 2*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/2*d*x +
1/2*c)^2 + a)^(9/2))/d

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maple [A]  time = 1.02, size = 67, normalized size = 0.73 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (35 \left (\sin ^{2}\left (d x +c \right )\right )+20 \sin \left (d x +c \right )+8\right )}{315 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+20*sin(d*x+c)+8)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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