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

Optimal. Leaf size=116 \[ \frac {152 a^2 \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}-\frac {4 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {38 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d} \]

[Out]

-4/35*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/a/d+152/105*a^2*sin(d*x+c)/d/(
a+a*cos(d*x+c))^(1/2)+38/105*a*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.14, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2759, 2751, 2647, 2646} \[ \frac {152 a^2 \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d}-\frac {4 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {38 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(152*a^2*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (38*a*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(105*d)
- (4*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*a*d)

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]

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

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 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)]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \, dx &=\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {2 \int \left (\frac {5 a}{2}-a \cos (c+d x)\right ) (a+a \cos (c+d x))^{3/2} \, dx}{7 a}\\ &=-\frac {4 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {19}{35} \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac {38 a \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}-\frac {4 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {1}{105} (76 a) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {152 a^2 \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {38 a \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 d}-\frac {4 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 81, normalized size = 0.70 \[ \frac {a \left (735 \sin \left (\frac {1}{2} (c+d x)\right )+175 \sin \left (\frac {3}{2} (c+d x)\right )+63 \sin \left (\frac {5}{2} (c+d x)\right )+15 \sin \left (\frac {7}{2} (c+d x)\right )\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)}}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(735*Sin[(c + d*x)/2] + 175*Sin[(3*(c + d*x))/2] + 63*Sin[(5*(c
+ d*x))/2] + 15*Sin[(7*(c + d*x))/2]))/(420*d)

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fricas [A]  time = 0.99, size = 67, normalized size = 0.58 \[ \frac {2 \, {\left (15 \, a \cos \left (d x + c\right )^{3} + 39 \, a \cos \left (d x + c\right )^{2} + 52 \, a \cos \left (d x + c\right ) + 104 \, a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a*cos(d*x + c)^3 + 39*a*cos(d*x + c)^2 + 52*a*cos(d*x + c) + 104*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x
+ c)/(d*cos(d*x + c) + d)

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giac [A]  time = 0.64, size = 109, normalized size = 0.94 \[ \frac {1}{420} \, \sqrt {2} {\left (\frac {15 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {63 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {175 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {735 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*sqrt(2)*(15*a*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c)/d + 63*a*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*
d*x + 5/2*c)/d + 175*a*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c)/d + 735*a*sgn(cos(1/2*d*x + 1/2*c))*sin(
1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.16, size = 86, normalized size = 0.74 \[ \frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (60 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38\right ) \sqrt {2}}{105 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(60*cos(1/2*d*x+1/2*c)^6-12*cos(1/2*d*x+1/2*c)^4+19*cos(1/2*d*
x+1/2*c)^2+38)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 1.01, size = 69, normalized size = 0.59 \[ \frac {{\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(15*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 63*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 175*sqrt(2)*a*sin(3/2*d*x + 3/2
*c) + 735*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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