3.142 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=191 \[ -\frac{16384 a^6 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac{4096 a^5 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac{512 a^4 \cos ^5(c+d x)}{1287 d \sqrt{a \sin (c+d x)+a}}-\frac{128 a^3 \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{429 d}-\frac{8 a^2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d} \]

[Out]

(-16384*a^6*Cos[c + d*x]^5)/(45045*d*(a + a*Sin[c + d*x])^(5/2)) - (4096*a^5*Cos[c + d*x]^5)/(9009*d*(a + a*Si
n[c + d*x])^(3/2)) - (512*a^4*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) - (128*a^3*Cos[c + d*x]^5*Sqrt
[a + a*Sin[c + d*x]])/(429*d) - (8*a^2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(39*d) - (2*a*Cos[c + d*x]^5
*(a + a*Sin[c + d*x])^(5/2))/(15*d)

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Rubi [A]  time = 0.365284, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac{16384 a^6 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac{4096 a^5 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac{512 a^4 \cos ^5(c+d x)}{1287 d \sqrt{a \sin (c+d x)+a}}-\frac{128 a^3 \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{429 d}-\frac{8 a^2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-16384*a^6*Cos[c + d*x]^5)/(45045*d*(a + a*Sin[c + d*x])^(5/2)) - (4096*a^5*Cos[c + d*x]^5)/(9009*d*(a + a*Si
n[c + d*x])^(3/2)) - (512*a^4*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) - (128*a^3*Cos[c + d*x]^5*Sqrt
[a + a*Sin[c + d*x]])/(429*d) - (8*a^2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(39*d) - (2*a*Cos[c + d*x]^5
*(a + a*Sin[c + d*x])^(5/2))/(15*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 ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac{1}{3} (4 a) \int \cos ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac{8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac{1}{39} \left (64 a^2\right ) \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{128 a^3 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}-\frac{8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac{1}{143} \left (256 a^3\right ) \int \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{512 a^4 \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}-\frac{128 a^3 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}-\frac{8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac{\left (2048 a^4\right ) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{1287}\\ &=-\frac{4096 a^5 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac{512 a^4 \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}-\frac{128 a^3 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}-\frac{8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac{\left (8192 a^5\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{9009}\\ &=-\frac{16384 a^6 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac{4096 a^5 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac{512 a^4 \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}-\frac{128 a^3 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{429 d}-\frac{8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.23296, size = 92, normalized size = 0.48 \[ -\frac{2 a^3 \left (3003 \sin ^5(c+d x)+19635 \sin ^4(c+d x)+55230 \sin ^3(c+d x)+86870 \sin ^2(c+d x)+81815 \sin (c+d x)+41735\right ) \cos ^5(c+d x) \sqrt{a (\sin (c+d x)+1)}}{45045 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-2*a^3*Cos[c + d*x]^5*Sqrt[a*(1 + Sin[c + d*x])]*(41735 + 81815*Sin[c + d*x] + 86870*Sin[c + d*x]^2 + 55230*S
in[c + d*x]^3 + 19635*Sin[c + d*x]^4 + 3003*Sin[c + d*x]^5))/(45045*d*(1 + Sin[c + d*x])^3)

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Maple [A]  time = 0.116, size = 97, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{4} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 3003\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}+19635\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+55230\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+86870\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+81815\,\sin \left ( dx+c \right ) +41735 \right ) }{45045\,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)^4*(a+a*sin(d*x+c))^(7/2),x)

[Out]

2/45045*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)^3*(3003*sin(d*x+c)^5+19635*sin(d*x+c)^4+55230*sin(d*x+c)^3+86870*sin
(d*x+c)^2+81815*sin(d*x+c)+41735)/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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

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Fricas [A]  time = 1.69027, size = 683, normalized size = 3.58 \begin{align*} \frac{2 \,{\left (3003 \, a^{3} \cos \left (d x + c\right )^{8} + 13629 \, a^{3} \cos \left (d x + c\right )^{7} - 17346 \, a^{3} \cos \left (d x + c\right )^{6} - 36932 \, a^{3} \cos \left (d x + c\right )^{5} + 1280 \, a^{3} \cos \left (d x + c\right )^{4} - 2048 \, a^{3} \cos \left (d x + c\right )^{3} + 4096 \, a^{3} \cos \left (d x + c\right )^{2} - 16384 \, a^{3} \cos \left (d x + c\right ) - 32768 \, a^{3} +{\left (3003 \, a^{3} \cos \left (d x + c\right )^{7} - 10626 \, a^{3} \cos \left (d x + c\right )^{6} - 27972 \, a^{3} \cos \left (d x + c\right )^{5} + 8960 \, a^{3} \cos \left (d x + c\right )^{4} + 10240 \, a^{3} \cos \left (d x + c\right )^{3} + 12288 \, a^{3} \cos \left (d x + c\right )^{2} + 16384 \, a^{3} \cos \left (d x + c\right ) + 32768 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{45045 \,{\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)^4*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/45045*(3003*a^3*cos(d*x + c)^8 + 13629*a^3*cos(d*x + c)^7 - 17346*a^3*cos(d*x + c)^6 - 36932*a^3*cos(d*x + c
)^5 + 1280*a^3*cos(d*x + c)^4 - 2048*a^3*cos(d*x + c)^3 + 4096*a^3*cos(d*x + c)^2 - 16384*a^3*cos(d*x + c) - 3
2768*a^3 + (3003*a^3*cos(d*x + c)^7 - 10626*a^3*cos(d*x + c)^6 - 27972*a^3*cos(d*x + c)^5 + 8960*a^3*cos(d*x +
c)^4 + 10240*a^3*cos(d*x + c)^3 + 12288*a^3*cos(d*x + c)^2 + 16384*a^3*cos(d*x + c) + 32768*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)**4*(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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)