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

Optimal. Leaf size=205 \[ -\frac{2 \sin ^4(c+d x) \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{11 a^2 d}-\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{385 a^3 d}+\frac{8 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{1155 a^2 d}+\frac{14 \sin ^4(c+d x) \cos (c+d x)}{33 a d \sqrt{a \sin (c+d x)+a}}-\frac{2 \sin ^3(c+d x) \cos (c+d x)}{231 a d \sqrt{a \sin (c+d x)+a}}-\frac{4 \cos (c+d x)}{165 a d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-4*Cos[c + d*x])/(165*a*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^3)/(231*a*d*Sqrt[a + a*Sin
[c + d*x]]) + (14*Cos[c + d*x]*Sin[c + d*x]^4)/(33*a*d*Sqrt[a + a*Sin[c + d*x]]) + (8*Cos[c + d*x]*Sqrt[a + a*
Sin[c + d*x]])/(1155*a^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*a^2*d) - (4*Cos[c +
d*x]*(a + a*Sin[c + d*x])^(3/2))/(385*a^3*d)

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Rubi [A]  time = 0.776041, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {2880, 2770, 2759, 2751, 2646, 3046, 2981} \[ -\frac{2 \sin ^4(c+d x) \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{11 a^2 d}-\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{385 a^3 d}+\frac{8 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{1155 a^2 d}+\frac{14 \sin ^4(c+d x) \cos (c+d x)}{33 a d \sqrt{a \sin (c+d x)+a}}-\frac{2 \sin ^3(c+d x) \cos (c+d x)}{231 a d \sqrt{a \sin (c+d x)+a}}-\frac{4 \cos (c+d x)}{165 a d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-4*Cos[c + d*x])/(165*a*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^3)/(231*a*d*Sqrt[a + a*Sin
[c + d*x]]) + (14*Cos[c + d*x]*Sin[c + d*x]^4)/(33*a*d*Sqrt[a + a*Sin[c + d*x]]) + (8*Cos[c + d*x]*Sqrt[a + a*
Sin[c + d*x]])/(1155*a^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*a^2*d) - (4*Cos[c +
d*x]*(a + a*Sin[c + d*x])^(3/2))/(385*a^3*d)

Rule 2880

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[-2/(a*b*d), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 2), x], x] + Dist[1/a^2
, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}
, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -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 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 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 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 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp
[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-
1)] && NeQ[m + n + 2, 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 \frac{\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{\int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \left (1+\sin ^2(c+d x)\right ) \, dx}{a^2}-\frac{2 \int \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{4 \cos (c+d x) \sin ^4(c+d x)}{9 a 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 a^2 d}+\frac{2 \int \sin ^3(c+d x) \left (\frac{19 a}{2}+\frac{1}{2} a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{11 a^3}-\frac{16 \int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{9 a^2}\\ &=\frac{32 \cos (c+d x) \sin ^3(c+d x)}{63 a d \sqrt{a+a \sin (c+d x)}}+\frac{14 \cos (c+d x) \sin ^4(c+d x)}{33 a 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 a^2 d}-\frac{32 \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{21 a^2}+\frac{179 \int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{99 a^2}\\ &=-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt{a+a \sin (c+d x)}}+\frac{14 \cos (c+d x) \sin ^4(c+d x)}{33 a 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 a^2 d}+\frac{64 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^3 d}-\frac{64 \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{105 a^3}+\frac{358 \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{231 a^2}\\ &=-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt{a+a \sin (c+d x)}}+\frac{14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt{a+a \sin (c+d x)}}-\frac{128 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 a^2 d}-\frac{2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 a^2 d}-\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d}+\frac{716 \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{1155 a^3}-\frac{32 \int \sqrt{a+a \sin (c+d x)} \, dx}{45 a^2}\\ &=\frac{64 \cos (c+d x)}{45 a d \sqrt{a+a \sin (c+d x)}}-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt{a+a \sin (c+d x)}}+\frac{14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt{a+a \sin (c+d x)}}+\frac{8 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{1155 a^2 d}-\frac{2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 a^2 d}-\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d}+\frac{358 \int \sqrt{a+a \sin (c+d x)} \, dx}{495 a^2}\\ &=-\frac{4 \cos (c+d x)}{165 a d \sqrt{a+a \sin (c+d x)}}-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt{a+a \sin (c+d x)}}+\frac{14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt{a+a \sin (c+d x)}}+\frac{8 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{1155 a^2 d}-\frac{2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 a^2 d}-\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d}\\ \end{align*}

Mathematica [A]  time = 5.40543, size = 102, normalized size = 0.5 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 (-475 \sin (c+d x)+105 \sin (3 (c+d x))+140 \cos (2 (c+d x))-204)}{2310 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(-204 + 140*Cos[2*(c + d*x)] - 475*Sin[c +
d*x] + 105*Sin[3*(c + d*x)]))/(2310*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.838, size = 77, normalized size = 0.4 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 105\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+70\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+40\,\sin \left ( dx+c \right ) +16 \right ) }{1155\,ad\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)^3/(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/1155/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(105*sin(d*x+c)^3+70*sin(d*x+c)^2+40*sin(d*x+c)+16)/cos(d*x+c)/(a+a*s
in(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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)^3/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^3/(a*sin(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 1.08678, size = 456, normalized size = 2.22 \begin{align*} -\frac{2 \,{\left (105 \, \cos \left (d x + c\right )^{6} - 140 \, \cos \left (d x + c\right )^{5} - 460 \, \cos \left (d x + c\right )^{4} + 274 \, \cos \left (d x + c\right )^{3} + 607 \, \cos \left (d x + c\right )^{2} +{\left (105 \, \cos \left (d x + c\right )^{5} + 245 \, \cos \left (d x + c\right )^{4} - 215 \, \cos \left (d x + c\right )^{3} - 489 \, \cos \left (d x + c\right )^{2} + 118 \, \cos \left (d x + c\right ) + 236\right )} \sin \left (d x + c\right ) - 118 \, \cos \left (d x + c\right ) - 236\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{1155 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} 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)^3/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2/1155*(105*cos(d*x + c)^6 - 140*cos(d*x + c)^5 - 460*cos(d*x + c)^4 + 274*cos(d*x + c)^3 + 607*cos(d*x + c)^
2 + (105*cos(d*x + c)^5 + 245*cos(d*x + c)^4 - 215*cos(d*x + c)^3 - 489*cos(d*x + c)^2 + 118*cos(d*x + c) + 23
6)*sin(d*x + c) - 118*cos(d*x + c) - 236)*sqrt(a*sin(d*x + c) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) +
a^2*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)**3/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 2.32672, size = 358, normalized size = 1.75 \begin{align*} -\frac{\frac{2 \,{\left ({\left ({\left ({\left ({\left ({\left ({\left (\frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{14}} + \frac{11 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{264 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{693 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{693 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{264 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{11 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{14}}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{11}{2}}} - \frac{59 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{39}{2}}}}{1182720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/1182720*(2*(((((((2*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^14 + 11*sgn(tan(1/2*d*x + 1/2*c)
+ 1)/a^14)*tan(1/2*d*x + 1/2*c)^2 - 264*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^14)*tan(1/2*d*x + 1/2*c) + 693*sgn(ta
n(1/2*d*x + 1/2*c) + 1)/a^14)*tan(1/2*d*x + 1/2*c) - 693*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^14)*tan(1/2*d*x + 1/2
*c) + 264*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^14)*tan(1/2*d*x + 1/2*c)^2 - 11*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^14)*
tan(1/2*d*x + 1/2*c)^2 - 2*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^14)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(11/2) - 59*sqrt
(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(39/2))/d