3.144 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=159 \[ -\frac{4096 a^5 \cos ^3(c+d x)}{3465 d (a \sin (c+d x)+a)^{3/2}}-\frac{1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt{a \sin (c+d x)+a}}-\frac{128 a^3 \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{231 d}-\frac{32 a^2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{99 d}-\frac{2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d} \]

[Out]

(-4096*a^5*Cos[c + d*x]^3)/(3465*d*(a + a*Sin[c + d*x])^(3/2)) - (1024*a^4*Cos[c + d*x]^3)/(1155*d*Sqrt[a + a*
Sin[c + d*x]]) - (128*a^3*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(231*d) - (32*a^2*Cos[c + d*x]^3*(a + a*Sin
[c + d*x])^(3/2))/(99*d) - (2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*d)

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Rubi [A]  time = 0.2921, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac{4096 a^5 \cos ^3(c+d x)}{3465 d (a \sin (c+d x)+a)^{3/2}}-\frac{1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt{a \sin (c+d x)+a}}-\frac{128 a^3 \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{231 d}-\frac{32 a^2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{99 d}-\frac{2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-4096*a^5*Cos[c + d*x]^3)/(3465*d*(a + a*Sin[c + d*x])^(3/2)) - (1024*a^4*Cos[c + d*x]^3)/(1155*d*Sqrt[a + a*
Sin[c + d*x]]) - (128*a^3*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(231*d) - (32*a^2*Cos[c + d*x]^3*(a + a*Sin
[c + d*x])^(3/2))/(99*d) - (2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*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))^{7/2} \, dx &=-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac{1}{11} (16 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac{32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac{1}{33} \left (64 a^2\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{128 a^3 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{231 d}-\frac{32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac{1}{231} \left (512 a^3\right ) \int \cos ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt{a+a \sin (c+d x)}}-\frac{128 a^3 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{231 d}-\frac{32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac{\left (2048 a^4\right ) \int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{1155}\\ &=-\frac{4096 a^5 \cos ^3(c+d x)}{3465 d (a+a \sin (c+d x))^{3/2}}-\frac{1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt{a+a \sin (c+d x)}}-\frac{128 a^3 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{231 d}-\frac{32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac{2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}\\ \end{align*}

Mathematica [A]  time = 0.204262, size = 82, normalized size = 0.52 \[ -\frac{2 a^3 \left (315 \sin ^4(c+d x)+1820 \sin ^3(c+d x)+4530 \sin ^2(c+d x)+6396 \sin (c+d x)+5419\right ) \cos ^3(c+d x) \sqrt{a (\sin (c+d x)+1)}}{3465 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-2*a^3*Cos[c + d*x]^3*Sqrt[a*(1 + Sin[c + d*x])]*(5419 + 6396*Sin[c + d*x] + 4530*Sin[c + d*x]^2 + 1820*Sin[c
+ d*x]^3 + 315*Sin[c + d*x]^4))/(3465*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.11, size = 87, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{4} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2} \left ( 315\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+1820\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+4530\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+6396\,\sin \left ( dx+c \right ) +5419 \right ) }{3465\,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))^(7/2),x)

[Out]

-2/3465*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)^2*(315*sin(d*x+c)^4+1820*sin(d*x+c)^3+4530*sin(d*x+c)^2+6396*sin(d*x
+c)+5419)/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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(7/2)*cos(d*x + c)^2, x)

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Fricas [A]  time = 1.65499, size = 522, normalized size = 3.28 \begin{align*} \frac{2 \,{\left (315 \, a^{3} \cos \left (d x + c\right )^{6} + 1505 \, a^{3} \cos \left (d x + c\right )^{5} - 2150 \, a^{3} \cos \left (d x + c\right )^{4} - 4876 \, a^{3} \cos \left (d x + c\right )^{3} + 512 \, a^{3} \cos \left (d x + c\right )^{2} - 2048 \, a^{3} \cos \left (d x + c\right ) - 4096 \, a^{3} +{\left (315 \, a^{3} \cos \left (d x + c\right )^{5} - 1190 \, a^{3} \cos \left (d x + c\right )^{4} - 3340 \, a^{3} \cos \left (d x + c\right )^{3} + 1536 \, a^{3} \cos \left (d x + c\right )^{2} + 2048 \, a^{3} \cos \left (d x + c\right ) + 4096 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{3465 \,{\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))^(7/2),x, algorithm="fricas")

[Out]

2/3465*(315*a^3*cos(d*x + c)^6 + 1505*a^3*cos(d*x + c)^5 - 2150*a^3*cos(d*x + c)^4 - 4876*a^3*cos(d*x + c)^3 +
512*a^3*cos(d*x + c)^2 - 2048*a^3*cos(d*x + c) - 4096*a^3 + (315*a^3*cos(d*x + c)^5 - 1190*a^3*cos(d*x + c)^4
- 3340*a^3*cos(d*x + c)^3 + 1536*a^3*cos(d*x + c)^2 + 2048*a^3*cos(d*x + c) + 4096*a^3)*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(cos(d*x+c)**2*(a+a*sin(d*x+c))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(7/2)*cos(d*x + c)^2, x)