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

[Out]

-256/6435*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-64/1287*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)+4/39*cos
(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/d-2/15*cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2)/a/d-56/1287*a^2*cos(d*x+c)^5/d/(a+
a*sin(d*x+c))^(1/2)-14/429*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.51, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 9.08, 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]

-1/51480*(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)]))/(d*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.46, size = 210, normalized size = 1.12 \[ \frac {2 \, {\left (429 \, a \cos \left (d x + c\right )^{8} + 957 \, a \cos \left (d x + c\right )^{7} - 633 \, a \cos \left (d x + c\right )^{6} - 1301 \, a \cos \left (d x + c\right )^{5} + 20 \, a \cos \left (d x + c\right )^{4} - 32 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) + {\left (429 \, a \cos \left (d x + c\right )^{7} - 528 \, a \cos \left (d x + c\right )^{6} - 1161 \, a \cos \left (d x + c\right )^{5} + 140 \, a \cos \left (d x + c\right )^{4} + 160 \, a \cos \left (d x + c\right )^{3} + 192 \, a \cos \left (d x + c\right )^{2} + 256 \, a \cos \left (d x + c\right ) + 512 \, a\right )} \sin \left (d x + c\right ) - 512 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6435 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

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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giac [B]  time = 1.43, size = 474, normalized size = 2.52 \[ \text {result too large to display} \]

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]

1/2882880*sqrt(2)*(3465*a*cos(1/4*pi + 13/2*d*x + 13/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 5005*a*cos(1
/4*pi + 9/2*d*x + 9/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 27027*a*cos(1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c))/d - 135135*a*cos(1/4*pi + 1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
+ 3003*a*cos(-1/4*pi + 15/2*d*x + 15/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 4095*a*cos(-1/4*pi + 11/2*d*
x + 11/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 19305*a*cos(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi + 1
/2*d*x + 1/2*c))/d - 45045*a*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 8190*a*sgn
(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 11/2*d*x + 11/2*c)/d - 25740*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c
))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d + 30030*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 3/2*d*x + 3/2*c)
/d - 6930*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 13/2*d*x + 13/2*c)/d - 20020*a*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*x + 9/2*c)/d + 18018*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi
+ 5/2*d*x + 5/2*c)/d + 360360*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 1.11, size = 97, normalized size = 0.52 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{3} \left (429 \left (\sin ^{5}\left (d x +c \right )\right )+1815 \left (\sin ^{4}\left (d x +c \right )\right )+3075 \left (\sin ^{3}\left (d x +c \right )\right )+2765 \left (\sin ^{2}\left (d x +c \right )\right )+1580 \sin \left (d x +c \right )+632\right )}{6435 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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]

2/6435*(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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}\,{d x} \]

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

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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