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

Optimal. Leaf size=193 \[ \frac {2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}+\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {38 a \sin ^3(c+d x) \cos (c+d x)}{693 d \sqrt {a \sin (c+d x)+a}}-\frac {76 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{1155 a d}+\frac {152 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3465 d}-\frac {76 a \cos (c+d x)}{495 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-76/1155*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-76/495*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-38/693*a*cos(d*x+c
)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+2/99*a*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+152/3465*cos(d
*x+c)*(a+a*sin(d*x+c))^(1/2)/d+2/11*cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.57, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2879, 2976, 2981, 2770, 2759, 2751, 2646} \[ \frac {2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}+\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {38 a \sin ^3(c+d x) \cos (c+d x)}{693 d \sqrt {a \sin (c+d x)+a}}-\frac {76 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{1155 a d}+\frac {152 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3465 d}-\frac {76 a \cos (c+d x)}{495 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sin[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-76*a*Cos[c + d*x])/(495*d*Sqrt[a + a*Sin[c + d*x]]) - (38*a*Cos[c + d*x]*Sin[c + d*x]^3)/(693*d*Sqrt[a + a*S
in[c + d*x]]) + (2*a*Cos[c + d*x]*Sin[c + d*x]^4)/(99*d*Sqrt[a + a*Sin[c + d*x]]) + (152*Cos[c + d*x]*Sqrt[a +
a*Sin[c + d*x]])/(3465*d) + (2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) - (76*Cos[c + d*x
]*(a + a*Sin[c + d*x])^(3/2))/(1155*a*d)

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2759

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*
Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2879

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n]

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\int \sin ^3(c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{3/2} \, dx}{a^2}\\ &=\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {2 \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {3 a^2}{2}-\frac {1}{2} a^2 \sin (c+d x)\right ) \, dx}{11 a^2}\\ &=\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {19}{99} \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {38}{231} \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d}+\frac {76 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{1155 a}\\ &=-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {152 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3465 d}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d}+\frac {38}{495} \int \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {76 a \cos (c+d x)}{495 d \sqrt {a+a \sin (c+d x)}}-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {152 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3465 d}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d}\\ \end {align*}

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Mathematica [A]  time = 1.23, size = 109, normalized size = 0.56 \[ -\frac {\sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (7638 \sin (c+d x)-1330 \sin (3 (c+d x))-3540 \cos (2 (c+d x))+315 \cos (4 (c+d x))+5657)}{13860 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-1/13860*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(5657 - 3540*Cos[2*(c + d*x)] + 3
15*Cos[4*(c + d*x)] + 7638*Sin[c + d*x] - 1330*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.53, size = 151, normalized size = 0.78 \[ \frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} + 350 \, \cos \left (d x + c\right )^{5} - 500 \, \cos \left (d x + c\right )^{4} - 586 \, \cos \left (d x + c\right )^{3} + 17 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} - 35 \, \cos \left (d x + c\right )^{4} - 535 \, \cos \left (d x + c\right )^{3} + 51 \, \cos \left (d x + c\right )^{2} + 68 \, \cos \left (d x + c\right ) + 136\right )} \sin \left (d x + c\right ) - 68 \, \cos \left (d x + c\right ) - 136\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3465*(315*cos(d*x + c)^6 + 350*cos(d*x + c)^5 - 500*cos(d*x + c)^4 - 586*cos(d*x + c)^3 + 17*cos(d*x + c)^2
+ (315*cos(d*x + c)^5 - 35*cos(d*x + c)^4 - 535*cos(d*x + c)^3 + 51*cos(d*x + c)^2 + 68*cos(d*x + c) + 136)*si
n(d*x + c) - 68*cos(d*x + c) - 136)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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giac [A]  time = 0.26, size = 189, normalized size = 0.98 \[ \frac {1}{55440} \, \sqrt {2} \sqrt {a} {\left (\frac {385 \, \cos \left (\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {693 \, \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {6930 \, \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {315 \, \cos \left (-\frac {1}{4} \, \pi + \frac {11}{2} \, d x + \frac {11}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {495 \, \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {2310 \, \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/55440*sqrt(2)*sqrt(a)*(385*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*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*cos(1/4*pi + 1/2*d*x + 1/2*c)*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))/d + 315*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*c
os(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 2310*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(
cos(-1/4*pi + 1/2*d*x + 1/2*c))/d)

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maple [A]  time = 1.04, size = 85, normalized size = 0.44 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{2} \left (315 \left (\sin ^{4}\left (d x +c \right )\right )+665 \left (\sin ^{3}\left (d x +c \right )\right )+570 \left (\sin ^{2}\left (d x +c \right )\right )+456 \sin \left (d x +c \right )+304\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x)

[Out]

-2/3465*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^2*(315*sin(d*x+c)^4+665*sin(d*x+c)^3+570*sin(d*x+c)^2+456*sin(d*x+c)+3
04)/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^2*sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2), 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)**3*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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