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

Optimal. Leaf size=162 \[ \frac{2 a^2 \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt{a \cos (c+d x)+a}}+\frac{34 a^2 \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}}+\frac{68 a^2 \sin (c+d x)}{45 d \sqrt{a \cos (c+d x)+a}}+\frac{68 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}-\frac{136 a \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d} \]

[Out]

(68*a^2*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (34*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(63*d*Sqrt[a + a*
Cos[c + d*x]]) + (2*a^2*Cos[c + d*x]^4*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d*x]]) - (136*a*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(315*d) + (68*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(105*d)

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Rubi [A]  time = 0.247446, antiderivative size = 162, 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, 21, 2770, 2759, 2751, 2646} \[ \frac{2 a^2 \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt{a \cos (c+d x)+a}}+\frac{34 a^2 \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}}+\frac{68 a^2 \sin (c+d x)}{45 d \sqrt{a \cos (c+d x)+a}}+\frac{68 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}-\frac{136 a \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(68*a^2*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (34*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(63*d*Sqrt[a + a*
Cos[c + d*x]]) + (2*a^2*Cos[c + d*x]^4*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d*x]]) - (136*a*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(315*d) + (68*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(105*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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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))^{3/2} \, dx &=\frac{2 a^2 \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{2}{9} \int \frac{\cos ^3(c+d x) \left (\frac{17 a^2}{2}+\frac{17}{2} a^2 \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 a^2 \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{1}{9} (17 a) \int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{34 a^2 \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{1}{21} (34 a) \int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{34 a^2 \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{68 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac{68}{105} \int \left (\frac{3 a}{2}-a \cos (c+d x)\right ) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{34 a^2 \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}-\frac{136 a \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{68 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac{1}{45} (34 a) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{68 a^2 \sin (c+d x)}{45 d \sqrt{a+a \cos (c+d x)}}+\frac{34 a^2 \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}-\frac{136 a \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{68 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}\\ \end{align*}

Mathematica [A]  time = 0.243161, size = 93, normalized size = 0.57 \[ \frac{a \left (3780 \sin \left (\frac{1}{2} (c+d x)\right )+1050 \sin \left (\frac{3}{2} (c+d x)\right )+378 \sin \left (\frac{5}{2} (c+d x)\right )+135 \sin \left (\frac{7}{2} (c+d x)\right )+35 \sin \left (\frac{9}{2} (c+d x)\right )\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)}}{2520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3780*Sin[(c + d*x)/2] + 1050*Sin[(3*(c + d*x))/2] + 378*Sin[(5
*(c + d*x))/2] + 135*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d)

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Maple [A]  time = 0.759, size = 99, normalized size = 0.6 \begin{align*}{\frac{4\,{a}^{2}\sqrt{2}}{315\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 280\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-220\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+114\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+47\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+94 \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)^(3/2),x)

[Out]

4/315*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(280*cos(1/2*d*x+1/2*c)^8-220*cos(1/2*d*x+1/2*c)^6+114*cos(1/2
*d*x+1/2*c)^4+47*cos(1/2*d*x+1/2*c)^2+94)*2^(1/2)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [A]  time = 2.0319, size = 113, normalized size = 0.7 \begin{align*} \frac{{\left (35 \, \sqrt{2} a \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 135 \, \sqrt{2} a \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 378 \, \sqrt{2} a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 1050 \, \sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3780 \, \sqrt{2} a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{2520 \, 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))^(3/2),x, algorithm="maxima")

[Out]

1/2520*(35*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 135*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 378*sqrt(2)*a*sin(5/2*d*x + 5
/2*c) + 1050*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 3780*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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Fricas [A]  time = 1.58266, size = 219, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} + 102 \, a \cos \left (d x + c\right )^{2} + 136 \, a \cos \left (d x + c\right ) + 272 \, a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \,{\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))^(3/2),x, algorithm="fricas")

[Out]

2/315*(35*a*cos(d*x + c)^4 + 85*a*cos(d*x + c)^3 + 102*a*cos(d*x + c)^2 + 136*a*cos(d*x + c) + 272*a)*sqrt(a*c
os(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))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

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