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

Optimal. Leaf size=116 \[ -\frac{152 a^2 \cos (c+d x)}{105 d \sqrt{a \sin (c+d x)+a}}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}-\frac{38 a \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{105 d} \]

[Out]

(-152*a^2*Cos[c + d*x])/(105*d*Sqrt[a + a*Sin[c + d*x]]) - (38*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(105*d
) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(35*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*a*d)

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Rubi [A]  time = 0.135378, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2759, 2751, 2647, 2646} \[ -\frac{152 a^2 \cos (c+d x)}{105 d \sqrt{a \sin (c+d x)+a}}-\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}-\frac{38 a \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-152*a^2*Cos[c + d*x])/(105*d*Sqrt[a + a*Sin[c + d*x]]) - (38*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(105*d
) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(35*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*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))^{3/2} \, dx &=-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac{2 \int \left (\frac{5 a}{2}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{7 a}\\ &=\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac{19}{35} \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{38 a \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{105 d}+\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac{1}{105} (76 a) \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{152 a^2 \cos (c+d x)}{105 d \sqrt{a+a \sin (c+d x)}}-\frac{38 a \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{105 d}+\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}\\ \end{align*}

Mathematica [A]  time = 0.352085, size = 141, normalized size = 1.22 \[ \frac{(a (\sin (c+d x)+1))^{3/2} \left (735 \sin \left (\frac{1}{2} (c+d x)\right )-175 \sin \left (\frac{3}{2} (c+d x)\right )-63 \sin \left (\frac{5}{2} (c+d x)\right )+15 \sin \left (\frac{7}{2} (c+d x)\right )-735 \cos \left (\frac{1}{2} (c+d x)\right )-175 \cos \left (\frac{3}{2} (c+d x)\right )+63 \cos \left (\frac{5}{2} (c+d x)\right )+15 \cos \left (\frac{7}{2} (c+d x)\right )\right )}{420 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(3/2)*(-735*Cos[(c + d*x)/2] - 175*Cos[(3*(c + d*x))/2] + 63*Cos[(5*(c + d*x))/2] + 15
*Cos[(7*(c + d*x))/2] + 735*Sin[(c + d*x)/2] - 175*Sin[(3*(c + d*x))/2] - 63*Sin[(5*(c + d*x))/2] + 15*Sin[(7*
(c + d*x))/2]))/(420*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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Maple [A]  time = 0.609, size = 75, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 15\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+39\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+52\,\sin \left ( dx+c \right ) +104 \right ) }{105\,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))^(3/2),x)

[Out]

2/105*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(15*sin(d*x+c)^3+39*sin(d*x+c)^2+52*sin(d*x+c)+104)/cos(d*x+c)/(a+a*si
n(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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51588, size = 338, normalized size = 2.91 \begin{align*} \frac{2 \,{\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} - 43 \, a \cos \left (d x + c\right )^{2} - 143 \, a \cos \left (d x + c\right ) +{\left (15 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 67 \, a \cos \left (d x + c\right ) + 76 \, a\right )} \sin \left (d x + c\right ) - 76 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{105 \,{\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))^(3/2),x, algorithm="fricas")

[Out]

2/105*(15*a*cos(d*x + c)^4 + 39*a*cos(d*x + c)^3 - 43*a*cos(d*x + c)^2 - 143*a*cos(d*x + c) + (15*a*cos(d*x +
c)^3 - 24*a*cos(d*x + c)^2 - 67*a*cos(d*x + c) + 76*a)*sin(d*x + c) - 76*a)*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))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

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