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

Optimal. Leaf size=156 \[ -\frac{8 a^2 \cos ^5(c+d x)}{143 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^5(c+d x)}{1001 d (a \sin (c+d x)+a)^{3/2}}-\frac{256 a^4 \cos ^5(c+d x)}{5005 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}-\frac{6 a \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{143 d} \]

[Out]

(-256*a^4*Cos[c + d*x]^5)/(5005*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(1001*d*(a + a*Sin[c +
d*x])^(3/2)) - (8*a^2*Cos[c + d*x]^5)/(143*d*Sqrt[a + a*Sin[c + d*x]]) - (6*a*Cos[c + d*x]^5*Sqrt[a + a*Sin[c
+ d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d)

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Rubi [A]  time = 0.324718, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2856, 2674, 2673} \[ -\frac{8 a^2 \cos ^5(c+d x)}{143 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^5(c+d x)}{1001 d (a \sin (c+d x)+a)^{3/2}}-\frac{256 a^4 \cos ^5(c+d x)}{5005 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}-\frac{6 a \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{143 d} \]

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]

(-256*a^4*Cos[c + d*x]^5)/(5005*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(1001*d*(a + a*Sin[c +
d*x])^(3/2)) - (8*a^2*Cos[c + d*x]^5)/(143*d*Sqrt[a + a*Sin[c + d*x]]) - (6*a*Cos[c + d*x]^5*Sqrt[a + a*Sin[c
+ d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d)

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 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 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]

Rubi steps

\begin{align*} \int \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) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{3}{13} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{1}{143} (36 a) \int \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{8 a^2 \cos ^5(c+d x)}{143 d \sqrt{a+a \sin (c+d x)}}-\frac{6 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{1}{143} \left (32 a^2\right ) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{64 a^3 \cos ^5(c+d x)}{1001 d (a+a \sin (c+d x))^{3/2}}-\frac{8 a^2 \cos ^5(c+d x)}{143 d \sqrt{a+a \sin (c+d x)}}-\frac{6 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{\left (128 a^3\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{1001}\\ &=-\frac{256 a^4 \cos ^5(c+d x)}{5005 d (a+a \sin (c+d x))^{5/2}}-\frac{64 a^3 \cos ^5(c+d x)}{1001 d (a+a \sin (c+d x))^{3/2}}-\frac{8 a^2 \cos ^5(c+d x)}{143 d \sqrt{a+a \sin (c+d x)}}-\frac{6 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}\\ \end{align*}

Mathematica [A]  time = 5.03024, size = 110, normalized size = 0.71 \[ -\frac{a \sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 (28230 \sin (c+d x)-3290 \sin (3 (c+d x))-12600 \cos (2 (c+d x))+385 \cos (4 (c+d x))+19559)}{20020 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

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]

-(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(19559 - 12600*Cos[2*(c + d*x)] + 385*C
os[4*(c + d*x)] + 28230*Sin[c + d*x] - 3290*Sin[3*(c + d*x)]))/(20020*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.777, size = 87, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 385\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+1645\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+2765\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+2295\,\sin \left ( dx+c \right ) +918 \right ) }{5005\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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]

2/5005*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(385*sin(d*x+c)^4+1645*sin(d*x+c)^3+2765*sin(d*x+c)^2+2295*sin(d*x+
c)+918)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\,{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+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.05441, size = 533, normalized size = 3.42 \begin{align*} \frac{2 \,{\left (385 \, a \cos \left (d x + c\right )^{7} - 490 \, a \cos \left (d x + c\right )^{6} - 1015 \, a \cos \left (d x + c\right )^{5} + 20 \, a \cos \left (d x + c\right )^{4} - 32 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) -{\left (385 \, a \cos \left (d x + c\right )^{6} + 875 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 160 \, a \cos \left (d x + c\right )^{3} - 192 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - 512 \, a\right )} \sin \left (d x + c\right ) - 512 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{5005 \,{\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\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+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/5005*(385*a*cos(d*x + c)^7 - 490*a*cos(d*x + c)^6 - 1015*a*cos(d*x + c)^5 + 20*a*cos(d*x + c)^4 - 32*a*cos(d
*x + c)^3 + 64*a*cos(d*x + c)^2 - 256*a*cos(d*x + c) - (385*a*cos(d*x + c)^6 + 875*a*cos(d*x + c)^5 - 140*a*co
s(d*x + c)^4 - 160*a*cos(d*x + c)^3 - 192*a*cos(d*x + c)^2 - 256*a*cos(d*x + c) - 512*a)*sin(d*x + c) - 512*a)
*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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

[Out]

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