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

Optimal. Leaf size=156 \[ -\frac {64 a^3 \cos ^3(c+d x)}{385 d (a \sin (c+d x)+a)^{3/2}}-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}+\frac {4 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{33 d}-\frac {6 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{77 d} \]

[Out]

-64/385*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)+4/33*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-2/11*cos(d*x+c)^3
*(a+a*sin(d*x+c))^(5/2)/a/d-48/385*a^2*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-6/77*a*cos(d*x+c)^3*(a+a*sin(d*x+
c))^(1/2)/d

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Rubi [A]  time = 0.43, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2878, 2856, 2674, 2673} \[ -\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a \sin (c+d x)+a}}-\frac {64 a^3 \cos ^3(c+d x)}{385 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}+\frac {4 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{33 d}-\frac {6 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{77 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x]^3)/(385*d*(a + a*Sin[c + d*x])^(3/2)) - (48*a^2*Cos[c + d*x]^3)/(385*d*Sqrt[a + a*Sin[c
+ d*x]]) - (6*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(77*d) + (4*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2)
)/(33*d) - (2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*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 ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {2 \int \cos ^2(c+d x) \left (\frac {5 a}{2}-3 a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{11 a}\\ &=\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {3}{11} \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {1}{77} (24 a) \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {1}{385} \left (96 a^2\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {64 a^3 \cos ^3(c+d x)}{385 d (a+a \sin (c+d x))^{3/2}}-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}\\ \end {align*}

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Mathematica [A]  time = 1.92, size = 110, normalized size = 0.71 \[ -\frac {a \sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (5076 \sin (c+d x)-700 \sin (3 (c+d x))-2280 \cos (2 (c+d x))+105 \cos (4 (c+d x))+4159)}{4620 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4620*(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(4159 - 2280*Cos[2*(c + d*x)] +
105*Cos[4*(c + d*x)] + 5076*Sin[c + d*x] - 700*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.49, size = 166, normalized size = 1.06 \[ \frac {2 \, {\left (105 \, a \cos \left (d x + c\right )^{6} + 245 \, a \cos \left (d x + c\right )^{5} - 185 \, a \cos \left (d x + c\right )^{4} - 397 \, a \cos \left (d x + c\right )^{3} + 24 \, a \cos \left (d x + c\right )^{2} - 96 \, a \cos \left (d x + c\right ) + {\left (105 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 325 \, a \cos \left (d x + c\right )^{3} + 72 \, a \cos \left (d x + c\right )^{2} + 96 \, a \cos \left (d x + c\right ) + 192 \, a\right )} \sin \left (d x + c\right ) - 192 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1155 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(105*a*cos(d*x + c)^6 + 245*a*cos(d*x + c)^5 - 185*a*cos(d*x + c)^4 - 397*a*cos(d*x + c)^3 + 24*a*cos(d
*x + c)^2 - 96*a*cos(d*x + c) + (105*a*cos(d*x + c)^5 - 140*a*cos(d*x + c)^4 - 325*a*cos(d*x + c)^3 + 72*a*cos
(d*x + c)^2 + 96*a*cos(d*x + c) + 192*a)*sin(d*x + c) - 192*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*si
n(d*x + c) + d)

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giac [B]  time = 0.50, size = 288, normalized size = 1.85 \[ \frac {1}{55440} \, \sqrt {2} {\left (\frac {385 \, a \cos \left (\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {693 \, a \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {6930 \, a \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {315 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {11}{2} \, d x + \frac {11}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {495 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {2310 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {990 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} - \frac {770 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {13860 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55440*sqrt(2)*(385*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 - 693*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 - 6930*a*cos(1/4*pi + 1/2*d*x + 1/2*c)*sgn(cos(-1/4*p
i + 1/2*d*x + 1/2*c))/d + 315*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 - 495*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 - 2310*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 - 990*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 - 770*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 + 13860*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 = 0.86, size = 87, normalized size = 0.56 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} \left (105 \left (\sin ^{4}\left (d x +c \right )\right )+350 \left (\sin ^{3}\left (d x +c \right )\right )+465 \left (\sin ^{2}\left (d x +c \right )\right )+372 \sin \left (d x +c \right )+248\right )}{1155 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(105*sin(d*x+c)^4+350*sin(d*x+c)^3+465*sin(d*x+c)^2+372*sin(d*x+c)
+248)/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*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)^2*sin(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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