3.112 \(\int \cos ^3(c+d x) (a+a \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=203 \[ \frac{2 a^2 \sin (c+d x) \cos ^4(c+d x) \sqrt{a \cos (c+d x)+a}}{11 d}+\frac{46 a^3 \sin (c+d x) \cos ^4(c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{710 a^3 \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}-\frac{568 a^2 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{693 d}+\frac{284 a^3 \sin (c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{284 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{231 d} \]

[Out]

(284*a^3*Sin[c + d*x])/(99*d*Sqrt[a + a*Cos[c + d*x]]) + (710*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(693*d*Sqrt[a +
a*Cos[c + d*x]]) + (46*a^3*Cos[c + d*x]^4*Sin[c + d*x])/(99*d*Sqrt[a + a*Cos[c + d*x]]) - (568*a^2*Sqrt[a + a
*Cos[c + d*x]]*Sin[c + d*x])/(693*d) + (2*a^2*Cos[c + d*x]^4*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11*d) + (
284*a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(231*d)

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Rubi [A]  time = 0.362981, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2763, 2981, 2770, 2759, 2751, 2646} \[ \frac{2 a^2 \sin (c+d x) \cos ^4(c+d x) \sqrt{a \cos (c+d x)+a}}{11 d}+\frac{46 a^3 \sin (c+d x) \cos ^4(c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{710 a^3 \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}-\frac{568 a^2 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{693 d}+\frac{284 a^3 \sin (c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{284 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{231 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(284*a^3*Sin[c + d*x])/(99*d*Sqrt[a + a*Cos[c + d*x]]) + (710*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(693*d*Sqrt[a +
a*Cos[c + d*x]]) + (46*a^3*Cos[c + d*x]^4*Sin[c + d*x])/(99*d*Sqrt[a + a*Cos[c + d*x]]) - (568*a^2*Sqrt[a + a
*Cos[c + d*x]]*Sin[c + d*x])/(693*d) + (2*a^2*Cos[c + d*x]^4*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11*d) + (
284*a*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(231*d)

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

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 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 \cos ^3(c+d x) (a+a \cos (c+d x))^{5/2} \, dx &=\frac{2 a^2 \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{2}{11} \int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \left (\frac{19 a^2}{2}+\frac{23}{2} a^2 \cos (c+d x)\right ) \, dx\\ &=\frac{46 a^3 \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{1}{99} \left (355 a^2\right ) \int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{710 a^3 \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{46 a^3 \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{1}{231} \left (710 a^2\right ) \int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{710 a^3 \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{46 a^3 \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{284 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}+\frac{1}{231} (284 a) \int \left (\frac{3 a}{2}-a \cos (c+d x)\right ) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{710 a^3 \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{46 a^3 \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}-\frac{568 a^2 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{693 d}+\frac{2 a^2 \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{284 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}+\frac{1}{99} \left (142 a^2\right ) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{284 a^3 \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{710 a^3 \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{46 a^3 \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}-\frac{568 a^2 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{693 d}+\frac{2 a^2 \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{284 a (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{231 d}\\ \end{align*}

Mathematica [A]  time = 0.480828, size = 107, normalized size = 0.53 \[ \frac{a^2 \left (31878 \sin \left (\frac{1}{2} (c+d x)\right )+8778 \sin \left (\frac{3}{2} (c+d x)\right )+3465 \sin \left (\frac{5}{2} (c+d x)\right )+1287 \sin \left (\frac{7}{2} (c+d x)\right )+385 \sin \left (\frac{9}{2} (c+d x)\right )+63 \sin \left (\frac{11}{2} (c+d x)\right )\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)}}{11088 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(31878*Sin[(c + d*x)/2] + 8778*Sin[(3*(c + d*x))/2] + 3465*Si
n[(5*(c + d*x))/2] + 1287*Sin[(7*(c + d*x))/2] + 385*Sin[(9*(c + d*x))/2] + 63*Sin[(11*(c + d*x))/2]))/(11088*
d)

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Maple [A]  time = 0.794, size = 112, normalized size = 0.6 \begin{align*}{\frac{8\,{a}^{3}\sqrt{2}}{693\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 504\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}-364\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+178\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+75\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+100\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+200 \right ){\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+cos(d*x+c)*a)^(5/2),x)

[Out]

8/693*cos(1/2*d*x+1/2*c)*a^3*sin(1/2*d*x+1/2*c)*(504*cos(1/2*d*x+1/2*c)^10-364*cos(1/2*d*x+1/2*c)^8+178*cos(1/
2*d*x+1/2*c)^6+75*cos(1/2*d*x+1/2*c)^4+100*cos(1/2*d*x+1/2*c)^2+200)*2^(1/2)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [A]  time = 1.90858, size = 150, normalized size = 0.74 \begin{align*} \frac{{\left (63 \, \sqrt{2} a^{2} \sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ) + 385 \, \sqrt{2} a^{2} \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 1287 \, \sqrt{2} a^{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 3465 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 8778 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 31878 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{11088 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/11088*(63*sqrt(2)*a^2*sin(11/2*d*x + 11/2*c) + 385*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 1287*sqrt(2)*a^2*sin(7
/2*d*x + 7/2*c) + 3465*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 8778*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 31878*sqrt(2
)*a^2*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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Fricas [A]  time = 1.56357, size = 269, normalized size = 1.33 \begin{align*} \frac{2 \,{\left (63 \, a^{2} \cos \left (d x + c\right )^{5} + 224 \, a^{2} \cos \left (d x + c\right )^{4} + 355 \, a^{2} \cos \left (d x + c\right )^{3} + 426 \, a^{2} \cos \left (d x + c\right )^{2} + 568 \, a^{2} \cos \left (d x + c\right ) + 1136 \, a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{693 \,{\left (d \cos \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)^3*(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/693*(63*a^2*cos(d*x + c)^5 + 224*a^2*cos(d*x + c)^4 + 355*a^2*cos(d*x + c)^3 + 426*a^2*cos(d*x + c)^2 + 568*
a^2*cos(d*x + c) + 1136*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(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)**3*(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^3, x)