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

Optimal. Leaf size=188 \[ -\frac{56 a^2 \cos ^5(c+d x)}{1287 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^5(c+d x)}{1287 d (a \sin (c+d x)+a)^{3/2}}-\frac{256 a^4 \cos ^5(c+d x)}{6435 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}+\frac{4 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac{14 a \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{429 d} \]

[Out]

(-256*a^4*Cos[c + d*x]^5)/(6435*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(1287*d*(a + a*Sin[c +
d*x])^(3/2)) - (56*a^2*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) - (14*a*Cos[c + d*x]^5*Sqrt[a + a*Si
n[c + d*x]])/(429*d) + (4*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(39*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c +
d*x])^(5/2))/(15*a*d)

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Rubi [A]  time = 0.51479, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2878, 2856, 2674, 2673} \[ -\frac{56 a^2 \cos ^5(c+d x)}{1287 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^5(c+d x)}{1287 d (a \sin (c+d x)+a)^{3/2}}-\frac{256 a^4 \cos ^5(c+d x)}{6435 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}+\frac{4 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac{14 a \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{429 d} \]

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]

(-256*a^4*Cos[c + d*x]^5)/(6435*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(1287*d*(a + a*Sin[c +
d*x])^(3/2)) - (56*a^2*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) - (14*a*Cos[c + d*x]^5*Sqrt[a + a*Si
n[c + d*x]])/(429*d) + (4*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(39*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c +
d*x])^(5/2))/(15*a*d)

Rule 2878

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*g*(m + p + 2)), x] + Dist[1/
(b*(m + p + 2)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*(p + 1)*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 2, 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 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 ^2(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))^{5/2}}{15 a d}+\frac{2 \int \cos ^4(c+d x) \left (\frac{5 a}{2}-5 a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{15 a}\\ &=\frac{4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d}+\frac{7}{39} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{14 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}+\frac{4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d}+\frac{1}{143} (28 a) \int \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{56 a^2 \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}-\frac{14 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}+\frac{4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d}+\frac{\left (224 a^2\right ) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{1287}\\ &=-\frac{64 a^3 \cos ^5(c+d x)}{1287 d (a+a \sin (c+d x))^{3/2}}-\frac{56 a^2 \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}-\frac{14 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}+\frac{4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d}+\frac{\left (128 a^3\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{1287}\\ &=-\frac{256 a^4 \cos ^5(c+d x)}{6435 d (a+a \sin (c+d x))^{5/2}}-\frac{64 a^3 \cos ^5(c+d x)}{1287 d (a+a \sin (c+d x))^{3/2}}-\frac{56 a^2 \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}-\frac{14 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}+\frac{4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d}\\ \end{align*}

Mathematica [A]  time = 8.84998, size = 120, normalized size = 0.64 \[ -\frac{a \sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 (66470 \sin (c+d x)-14445 \sin (3 (c+d x))+429 \sin (5 (c+d x))-36640 \cos (2 (c+d x))+3630 \cos (4 (c+d x))+43122)}{51480 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

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]

-(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(43122 - 36640*Cos[2*(c + d*x)] + 3630*
Cos[4*(c + d*x)] + 66470*Sin[c + d*x] - 14445*Sin[3*(c + d*x)] + 429*Sin[5*(c + d*x)]))/(51480*d*(Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.693, size = 97, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 429\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}+1815\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+3075\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+2765\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+1580\,\sin \left ( dx+c \right ) +632 \right ) }{6435\,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)^2*(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/6435*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(429*sin(d*x+c)^5+1815*sin(d*x+c)^4+3075*sin(d*x+c)^3+2765*sin(d*x+
c)^2+1580*sin(d*x+c)+632)/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 )^{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)^2*(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)^2, x)

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Fricas [A]  time = 1.15855, size = 597, normalized size = 3.18 \begin{align*} \frac{2 \,{\left (429 \, a \cos \left (d x + c\right )^{8} + 957 \, a \cos \left (d x + c\right )^{7} - 633 \, a \cos \left (d x + c\right )^{6} - 1301 \, 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 (429 \, a \cos \left (d x + c\right )^{7} - 528 \, a \cos \left (d x + c\right )^{6} - 1161 \, 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}}{6435 \,{\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)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/6435*(429*a*cos(d*x + c)^8 + 957*a*cos(d*x + c)^7 - 633*a*cos(d*x + c)^6 - 1301*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) + (429*a*cos(d*x + c)^7 - 528*a*co
s(d*x + c)^6 - 1161*a*cos(d*x + c)^5 + 140*a*cos(d*x + c)^4 + 160*a*cos(d*x + c)^3 + 192*a*cos(d*x + c)^2 + 25
6*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)**2*(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 )^{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)^2*(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)^2, x)