3.4.36 \(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [336]

Optimal. Leaf size=158 \[ -\frac {4 \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d} \]

[Out]

-4/105*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a^2/d-4/45*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/63*cos(d*x+c)*sin(d*
x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+2/9*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+8/315*cos(d*x+c)*(a+a*sin
(d*x+c))^(1/2)/a/d

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Rubi [A]
time = 0.27, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2958, 3060, 2849, 2838, 2830, 2725} \begin {gather*} -\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 a^2 d}+\frac {2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}+\frac {8 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 a d}-\frac {4 \cos (c+d x)}{45 d \sqrt {a \sin (c+d x)+a}} \end {gather*}

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]

(-4*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin[c + d
*x]]) + (2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (8*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*
x]])/(315*a*d) - (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*a^2*d)

Rule 2725

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 2830

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*S
in[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 2838

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 2849

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 2958

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 3060

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)*Sqrt
[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 \frac {\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)) \sqrt {a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{9 a}\\ &=-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{21 a}\\ &=-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d}+\frac {4 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{105 a^2}\\ &=-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d}+\frac {2 \int \sqrt {a+a \sin (c+d x)} \, dx}{45 a}\\ &=-\frac {4 \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.90, size = 97, normalized size = 0.61 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (-124+60 \cos (2 (c+d x))-201 \sin (c+d x)+35 \sin (3 (c+d x)))}{630 d \sqrt {a (1+\sin (c+d x))}} \end {gather*}

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]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-124 + 60*Cos[2*(c + d*x)] - 2
01*Sin[c + d*x] + 35*Sin[3*(c + d*x)]))/(630*d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]
time = 5.10, size = 74, normalized size = 0.47

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{2} \left (35 \left (\sin ^{3}\left (d x +c \right )\right )+30 \left (\sin ^{2}\left (d x +c \right )\right )+24 \sin \left (d x +c \right )+16\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(74\)

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,method=_RETURNVERBOSE)

[Out]

-2/315*(1+sin(d*x+c))*(sin(d*x+c)-1)^2*(35*sin(d*x+c)^3+30*sin(d*x+c)^2+24*sin(d*x+c)+16)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 0.34, size = 136, normalized size = 0.86 \begin {gather*} \frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - 64 \, \cos \left (d x + c\right )^{3} - 82 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} - 69 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 26\right )} \sin \left (d x + c\right ) + 13 \, \cos \left (d x + c\right ) + 26\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \end {gather*}

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]

2/315*(35*cos(d*x + c)^5 + 40*cos(d*x + c)^4 - 64*cos(d*x + c)^3 - 82*cos(d*x + c)^2 - (35*cos(d*x + c)^4 - 5*
cos(d*x + c)^3 - 69*cos(d*x + c)^2 + 13*cos(d*x + c) + 26)*sin(d*x + c) + 13*cos(d*x + c) + 26)*sqrt(a*sin(d*x
+ c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [A]
time = 0.60, size = 103, normalized size = 0.65 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (280 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 540 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{315 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \end {gather*}

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]

-4/315*sqrt(2)*(280*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 540*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 +
378*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 105*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)/(a*d*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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 \end {gather*}

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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