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

Optimal. Leaf size=127 \[ -\frac{256 a^4 \cos ^3(c+d x)}{315 d (a \sin (c+d x)+a)^{3/2}}-\frac{64 a^3 \cos ^3(c+d x)}{105 d \sqrt{a \sin (c+d x)+a}}-\frac{8 a^2 \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{21 d}-\frac{2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d} \]

[Out]

(-256*a^4*Cos[c + d*x]^3)/(315*d*(a + a*Sin[c + d*x])^(3/2)) - (64*a^3*Cos[c + d*x]^3)/(105*d*Sqrt[a + a*Sin[c
+ d*x]]) - (8*a^2*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(21*d) - (2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^
(3/2))/(9*d)

________________________________________________________________________________________

Rubi [A]  time = 0.224175, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac{256 a^4 \cos ^3(c+d x)}{315 d (a \sin (c+d x)+a)^{3/2}}-\frac{64 a^3 \cos ^3(c+d x)}{105 d \sqrt{a \sin (c+d x)+a}}-\frac{8 a^2 \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{21 d}-\frac{2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*a^4*Cos[c + d*x]^3)/(315*d*(a + a*Sin[c + d*x])^(3/2)) - (64*a^3*Cos[c + d*x]^3)/(105*d*Sqrt[a + a*Sin[c
+ d*x]]) - (8*a^2*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(21*d) - (2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^
(3/2))/(9*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))^{5/2} \, dx &=-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac{1}{3} (4 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{8 a^2 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{21 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac{1}{21} \left (32 a^2\right ) \int \cos ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{64 a^3 \cos ^3(c+d x)}{105 d \sqrt{a+a \sin (c+d x)}}-\frac{8 a^2 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{21 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac{1}{105} \left (128 a^3\right ) \int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{256 a^4 \cos ^3(c+d x)}{315 d (a+a \sin (c+d x))^{3/2}}-\frac{64 a^3 \cos ^3(c+d x)}{105 d \sqrt{a+a \sin (c+d x)}}-\frac{8 a^2 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{21 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.264929, size = 69, normalized size = 0.54 \[ -\frac{2 \left (35 \sin ^3(c+d x)+165 \sin ^2(c+d x)+321 \sin (c+d x)+319\right ) \cos ^3(c+d x) (a (\sin (c+d x)+1))^{5/2}}{315 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^3*(a*(1 + Sin[c + d*x]))^(5/2)*(319 + 321*Sin[c + d*x] + 165*Sin[c + d*x]^2 + 35*Sin[c + d*x]
^3))/(315*d*(1 + Sin[c + d*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.106, size = 77, 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 ) ^{2} \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+165\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+321\,\sin \left ( dx+c \right ) +319 \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(cos(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)^2*(35*sin(d*x+c)^3+165*sin(d*x+c)^2+321*sin(d*x+c)+319)/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}} \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))^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.64077, size = 433, normalized size = 3.41 \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} - 226 \, a^{2} \cos \left (d x + c\right )^{3} + 32 \, a^{2} \cos \left (d x + c\right )^{2} - 128 \, a^{2} \cos \left (d x + c\right ) - 256 \, a^{2} -{\left (35 \, a^{2} \cos \left (d x + c\right )^{4} + 130 \, a^{2} \cos \left (d x + c\right )^{3} - 96 \, a^{2} \cos \left (d x + c\right )^{2} - 128 \, a^{2} \cos \left (d x + c\right ) - 256 \, 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(cos(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 - 226*a^2*cos(d*x + c)^3 + 32*a^2*cos(d*x + c)^2 - 128*a^
2*cos(d*x + c) - 256*a^2 - (35*a^2*cos(d*x + c)^4 + 130*a^2*cos(d*x + c)^3 - 96*a^2*cos(d*x + c)^2 - 128*a^2*c
os(d*x + c) - 256*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(cos(d*x+c)**2*(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}} \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))^(5/2),x, algorithm="giac")

[Out]

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