3.52 \(\int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=203 \[ -\frac{2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{11 d}-\frac{46 a^3 \sin ^4(c+d x) \cos (c+d x)}{99 d \sqrt{a \sin (c+d x)+a}}-\frac{710 a^3 \sin ^3(c+d x) \cos (c+d x)}{693 d \sqrt{a \sin (c+d x)+a}}+\frac{568 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{693 d}-\frac{284 a^3 \cos (c+d x)}{99 d \sqrt{a \sin (c+d x)+a}}-\frac{284 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{231 d} \]

[Out]

(-284*a^3*Cos[c + d*x])/(99*d*Sqrt[a + a*Sin[c + d*x]]) - (710*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(693*d*Sqrt[a
+ a*Sin[c + d*x]]) - (46*a^3*Cos[c + d*x]*Sin[c + d*x]^4)/(99*d*Sqrt[a + a*Sin[c + d*x]]) + (568*a^2*Cos[c + d
*x]*Sqrt[a + a*Sin[c + d*x]])/(693*d) - (2*a^2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) -
(284*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(231*d)

________________________________________________________________________________________

Rubi [A]  time = 0.351971, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2763, 2981, 2770, 2759, 2751, 2646} \[ -\frac{2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{11 d}-\frac{46 a^3 \sin ^4(c+d x) \cos (c+d x)}{99 d \sqrt{a \sin (c+d x)+a}}-\frac{710 a^3 \sin ^3(c+d x) \cos (c+d x)}{693 d \sqrt{a \sin (c+d x)+a}}+\frac{568 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{693 d}-\frac{284 a^3 \cos (c+d x)}{99 d \sqrt{a \sin (c+d x)+a}}-\frac{284 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{231 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-284*a^3*Cos[c + d*x])/(99*d*Sqrt[a + a*Sin[c + d*x]]) - (710*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(693*d*Sqrt[a
+ a*Sin[c + d*x]]) - (46*a^3*Cos[c + d*x]*Sin[c + d*x]^4)/(99*d*Sqrt[a + a*Sin[c + d*x]]) + (568*a^2*Cos[c + d
*x]*Sqrt[a + a*Sin[c + d*x]])/(693*d) - (2*a^2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) -
(284*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(231*d)

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && 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]

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]

Rubi steps

\begin{align*} \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}+\frac{2}{11} \int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \left (\frac{19 a^2}{2}+\frac{23}{2} a^2 \sin (c+d x)\right ) \, dx\\ &=-\frac{46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}+\frac{1}{99} \left (355 a^2\right ) \int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt{a+a \sin (c+d x)}}-\frac{46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}+\frac{1}{231} \left (710 a^2\right ) \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt{a+a \sin (c+d x)}}-\frac{46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}-\frac{284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d}+\frac{1}{231} (284 a) \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt{a+a \sin (c+d x)}}-\frac{46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt{a+a \sin (c+d x)}}+\frac{568 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{693 d}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}-\frac{284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d}+\frac{1}{99} \left (142 a^2\right ) \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{284 a^3 \cos (c+d x)}{99 d \sqrt{a+a \sin (c+d x)}}-\frac{710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt{a+a \sin (c+d x)}}-\frac{46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt{a+a \sin (c+d x)}}+\frac{568 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{693 d}-\frac{2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}-\frac{284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d}\\ \end{align*}

Mathematica [A]  time = 1.26886, size = 189, normalized size = 0.93 \[ -\frac{(a (\sin (c+d x)+1))^{5/2} \left (-31878 \sin \left (\frac{1}{2} (c+d x)\right )+8778 \sin \left (\frac{3}{2} (c+d x)\right )+3465 \sin \left (\frac{5}{2} (c+d x)\right )-1287 \sin \left (\frac{7}{2} (c+d x)\right )-385 \sin \left (\frac{9}{2} (c+d x)\right )+63 \sin \left (\frac{11}{2} (c+d x)\right )+31878 \cos \left (\frac{1}{2} (c+d x)\right )+8778 \cos \left (\frac{3}{2} (c+d x)\right )-3465 \cos \left (\frac{5}{2} (c+d x)\right )-1287 \cos \left (\frac{7}{2} (c+d x)\right )+385 \cos \left (\frac{9}{2} (c+d x)\right )+63 \cos \left (\frac{11}{2} (c+d x)\right )\right )}{11088 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-((a*(1 + Sin[c + d*x]))^(5/2)*(31878*Cos[(c + d*x)/2] + 8778*Cos[(3*(c + d*x))/2] - 3465*Cos[(5*(c + d*x))/2]
 - 1287*Cos[(7*(c + d*x))/2] + 385*Cos[(9*(c + d*x))/2] + 63*Cos[(11*(c + d*x))/2] - 31878*Sin[(c + d*x)/2] +
8778*Sin[(3*(c + d*x))/2] + 3465*Sin[(5*(c + d*x))/2] - 1287*Sin[(7*(c + d*x))/2] - 385*Sin[(9*(c + d*x))/2] +
 63*Sin[(11*(c + d*x))/2]))/(11088*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

________________________________________________________________________________________

Maple [A]  time = 0.628, size = 95, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 63\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}+224\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+355\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+426\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+568\,\sin \left ( dx+c \right ) +1136 \right ) }{693\,d\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(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/693*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(63*sin(d*x+c)^5+224*sin(d*x+c)^4+355*sin(d*x+c)^3+426*sin(d*x+c)^2+56
8*sin(d*x+c)+1136)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*sin(d*x + c)^3, x)

________________________________________________________________________________________

Fricas [A]  time = 1.53847, size = 508, normalized size = 2.5 \begin{align*} -\frac{2 \,{\left (63 \, a^{2} \cos \left (d x + c\right )^{6} + 224 \, a^{2} \cos \left (d x + c\right )^{5} - 320 \, a^{2} \cos \left (d x + c\right )^{4} - 874 \, a^{2} \cos \left (d x + c\right )^{3} + 593 \, a^{2} \cos \left (d x + c\right )^{2} + 1786 \, a^{2} \cos \left (d x + c\right ) + 800 \, a^{2} +{\left (63 \, a^{2} \cos \left (d x + c\right )^{5} - 161 \, a^{2} \cos \left (d x + c\right )^{4} - 481 \, a^{2} \cos \left (d x + c\right )^{3} + 393 \, a^{2} \cos \left (d x + c\right )^{2} + 986 \, a^{2} \cos \left (d x + c\right ) - 800 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{693 \,{\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/693*(63*a^2*cos(d*x + c)^6 + 224*a^2*cos(d*x + c)^5 - 320*a^2*cos(d*x + c)^4 - 874*a^2*cos(d*x + c)^3 + 593
*a^2*cos(d*x + c)^2 + 1786*a^2*cos(d*x + c) + 800*a^2 + (63*a^2*cos(d*x + c)^5 - 161*a^2*cos(d*x + c)^4 - 481*
a^2*cos(d*x + c)^3 + 393*a^2*cos(d*x + c)^2 + 986*a^2*cos(d*x + c) - 800*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c
) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

________________________________________________________________________________________

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(sin(d*x+c)**3*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*sin(d*x + c)^3, x)