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

Optimal. Leaf size=146 \[ -\frac{208 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 d}-\frac{832 a^3 \cos (c+d x)}{315 d \sqrt{a \sin (c+d x)+a}}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{7/2}}{9 a d}+\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{63 d}-\frac{26 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d} \]

[Out]

(-832*a^3*Cos[c + d*x])/(315*d*Sqrt[a + a*Sin[c + d*x]]) - (208*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(31
5*d) - (26*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*d) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(6
3*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/2))/(9*a*d)

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Rubi [A]  time = 0.154681, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2759, 2751, 2647, 2646} \[ -\frac{208 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 d}-\frac{832 a^3 \cos (c+d x)}{315 d \sqrt{a \sin (c+d x)+a}}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{7/2}}{9 a d}+\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{63 d}-\frac{26 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-832*a^3*Cos[c + d*x])/(315*d*Sqrt[a + a*Sin[c + d*x]]) - (208*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(31
5*d) - (26*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*d) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(6
3*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/2))/(9*a*d)

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 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

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 ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac{2 \int \left (\frac{7 a}{2}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{5/2} \, dx}{9 a}\\ &=\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac{13}{21} \int (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac{26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac{1}{105} (104 a) \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{208 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 d}-\frac{26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}+\frac{1}{315} \left (416 a^2\right ) \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{832 a^3 \cos (c+d x)}{315 d \sqrt{a+a \sin (c+d x)}}-\frac{208 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 d}-\frac{26 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d}+\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{63 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{7/2}}{9 a d}\\ \end{align*}

Mathematica [A]  time = 1.01727, size = 165, normalized size = 1.13 \[ \frac{(a (\sin (c+d x)+1))^{5/2} \left (8190 \sin \left (\frac{1}{2} (c+d x)\right )-2100 \sin \left (\frac{3}{2} (c+d x)\right )-756 \sin \left (\frac{5}{2} (c+d x)\right )+225 \sin \left (\frac{7}{2} (c+d x)\right )+35 \sin \left (\frac{9}{2} (c+d x)\right )-8190 \cos \left (\frac{1}{2} (c+d x)\right )-2100 \cos \left (\frac{3}{2} (c+d x)\right )+756 \cos \left (\frac{5}{2} (c+d x)\right )+225 \cos \left (\frac{7}{2} (c+d x)\right )-35 \cos \left (\frac{9}{2} (c+d x)\right )\right )}{2520 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(5/2)*(-8190*Cos[(c + d*x)/2] - 2100*Cos[(3*(c + d*x))/2] + 756*Cos[(5*(c + d*x))/2] +
225*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] + 8190*Sin[(c + d*x)/2] - 2100*Sin[(3*(c + d*x))/2] - 756*
Sin[(5*(c + d*x))/2] + 225*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^5)

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Maple [A]  time = 0.622, size = 85, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+130\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+219\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+292\,\sin \left ( dx+c \right ) +584 \right ) }{315\,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)^2*(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/315*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+130*sin(d*x+c)^3+219*sin(d*x+c)^2+292*sin(d*x+c)+584)
/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(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)^2, x)

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Fricas [A]  time = 1.37957, size = 437, normalized size = 2.99 \begin{align*} -\frac{2 \,{\left (35 \, a^{2} \cos \left (d x + c\right )^{5} - 95 \, a^{2} \cos \left (d x + c\right )^{4} - 289 \, a^{2} \cos \left (d x + c\right )^{3} + 263 \, a^{2} \cos \left (d x + c\right )^{2} + 838 \, a^{2} \cos \left (d x + c\right ) + 416 \, a^{2} -{\left (35 \, a^{2} \cos \left (d x + c\right )^{4} + 130 \, a^{2} \cos \left (d x + c\right )^{3} - 159 \, a^{2} \cos \left (d x + c\right )^{2} - 422 \, a^{2} \cos \left (d x + c\right ) + 416 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{315 \,{\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)^2*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/315*(35*a^2*cos(d*x + c)^5 - 95*a^2*cos(d*x + c)^4 - 289*a^2*cos(d*x + c)^3 + 263*a^2*cos(d*x + c)^2 + 838*
a^2*cos(d*x + c) + 416*a^2 - (35*a^2*cos(d*x + c)^4 + 130*a^2*cos(d*x + c)^3 - 159*a^2*cos(d*x + c)^2 - 422*a^
2*cos(d*x + c) + 416*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + 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(sin(d*x+c)**2*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(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)^2, x)