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

Optimal. Leaf size=60 \[ \frac {6 a \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

6/35*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-2/7*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)

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Rubi [A]  time = 0.15, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2856, 2673} \[ \frac {6 a \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))

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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac {2 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}-\frac {3}{7} \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac {6 a \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac {2 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.90, size = 82, normalized size = 1.37 \[ -\frac {2 (5 \sin (c+d x)+2) \sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}{35 a^2 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 121, normalized size = 2.02 \[ \frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{3} - 19 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} - 6 \, \cos \left (d x + c\right ) - 12\right )} \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right ) + 12\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*cos(d*x + c)^4 - 8*cos(d*x + c)^3 - 19*cos(d*x + c)^2 + (5*cos(d*x + c)^3 + 13*cos(d*x + c)^2 - 6*cos(
d*x + c) - 12)*sin(d*x + c) + 6*cos(d*x + c) + 12)*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 = 0.67, size = 216, normalized size = 3.60 \[ -\frac {4 \, {\left (\frac {6 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}}} - \frac {{\left ({\left ({\left ({\left (\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {14 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {35 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {35 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {14 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}\right )}}{35 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/35*(6*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(3/2) - (((((a^2*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2
*c) + 1) - 14*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 35*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*
tan(1/2*d*x + 1/2*c) - 35*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 14*a^2/sgn(tan(1/2*d*x + 1
/2*c) + 1))*tan(1/2*d*x + 1/2*c)^2 - a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/
d

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maple [A]  time = 0.87, size = 57, normalized size = 0.95 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (5 \sin \left (d x +c \right )+2\right )}{35 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(5*sin(d*x+c)+2)/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 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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