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

Optimal. Leaf size=95 \[ -\frac{16 a^2 \cos ^3(c+d x)}{35 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}}-\frac{2 a \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{7 d} \]

[Out]

(-64*a^3*Cos[c + d*x]^3)/(105*d*(a + a*Sin[c + d*x])^(3/2)) - (16*a^2*Cos[c + d*x]^3)/(35*d*Sqrt[a + a*Sin[c +
d*x]]) - (2*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(7*d)

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Rubi [A]  time = 0.167231, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac{16 a^2 \cos ^3(c+d x)}{35 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}}-\frac{2 a \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x]^3)/(105*d*(a + a*Sin[c + d*x])^(3/2)) - (16*a^2*Cos[c + d*x]^3)/(35*d*Sqrt[a + a*Sin[c +
d*x]]) - (2*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(7*d)

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac{2 a \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{7 d}+\frac{1}{7} (8 a) \int \cos ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{16 a^2 \cos ^3(c+d x)}{35 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{7 d}+\frac{1}{35} \left (32 a^2\right ) \int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{64 a^3 \cos ^3(c+d x)}{105 d (a+a \sin (c+d x))^{3/2}}-\frac{16 a^2 \cos ^3(c+d x)}{35 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.140159, size = 59, normalized size = 0.62 \[ -\frac{2 \left (15 \sin ^2(c+d x)+54 \sin (c+d x)+71\right ) \cos ^3(c+d x) (a (\sin (c+d x)+1))^{3/2}}{105 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^3*(a*(1 + Sin[c + d*x]))^(3/2)*(71 + 54*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(105*d*(1 + Sin[c
+ d*x])^3)

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Maple [A]  time = 0.113, size = 67, 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 ) ^{2} \left ( 15\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+54\,\sin \left ( dx+c \right ) +71 \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(cos(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)^2*(15*sin(d*x+c)^2+54*sin(d*x+c)+71)/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{3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)*cos(d*x + c)^2, x)

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Fricas [A]  time = 1.62546, size = 336, normalized size = 3.54 \begin{align*} -\frac{2 \,{\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} - 8 \, a \cos \left (d x + c\right )^{2} + 32 \, 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} - 32 \, a \cos \left (d x + c\right ) - 64 \, a\right )} \sin \left (d x + c\right ) + 64 \, 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(cos(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 - 8*a*cos(d*x + c)^2 + 32*a*cos(d*x + c) + (15*a*cos(d*x + c
)^3 - 24*a*cos(d*x + c)^2 - 32*a*cos(d*x + c) - 64*a)*sin(d*x + c) + 64*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(cos(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}} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)*cos(d*x + c)^2, x)