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

Optimal. Leaf size=127 \[ -\frac {256 a^4 \cos ^3(c+d x)}{315 d (a \sin (c+d x)+a)^{3/2}}-\frac {64 a^3 \cos ^3(c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {8 a^2 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d}-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d} \]

[Out]

-256/315*a^4*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/9*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-64/105*a^3*cos(
d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-8/21*a^2*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.22, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac {256 a^4 \cos ^3(c+d x)}{315 d (a \sin (c+d x)+a)^{3/2}}-\frac {64 a^3 \cos ^3(c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {8 a^2 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d}-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*a^4*Cos[c + d*x]^3)/(315*d*(a + a*Sin[c + d*x])^(3/2)) - (64*a^3*Cos[c + d*x]^3)/(105*d*Sqrt[a + a*Sin[c
+ d*x]]) - (8*a^2*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(21*d) - (2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^
(3/2))/(9*d)

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]

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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac {1}{3} (4 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {8 a^2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac {1}{21} \left (32 a^2\right ) \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {64 a^3 \cos ^3(c+d x)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {8 a^2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac {1}{105} \left (128 a^3\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {256 a^4 \cos ^3(c+d x)}{315 d (a+a \sin (c+d x))^{3/2}}-\frac {64 a^3 \cos ^3(c+d x)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {8 a^2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 69, normalized size = 0.54 \[ -\frac {2 \left (35 \sin ^3(c+d x)+165 \sin ^2(c+d x)+321 \sin (c+d x)+319\right ) \cos ^3(c+d x) (a (\sin (c+d x)+1))^{5/2}}{315 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^3*(a*(1 + Sin[c + d*x]))^(5/2)*(319 + 321*Sin[c + d*x] + 165*Sin[c + d*x]^2 + 35*Sin[c + d*x]
^3))/(315*d*(1 + Sin[c + d*x])^4)

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fricas [A]  time = 0.61, size = 167, normalized size = 1.31 \[ \frac {2 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{5} - 95 \, a^{2} \cos \left (d x + c\right )^{4} - 226 \, a^{2} \cos \left (d x + c\right )^{3} + 32 \, a^{2} \cos \left (d x + c\right )^{2} - 128 \, a^{2} \cos \left (d x + c\right ) - 256 \, a^{2} - {\left (35 \, a^{2} \cos \left (d x + c\right )^{4} + 130 \, a^{2} \cos \left (d x + c\right )^{3} - 96 \, a^{2} \cos \left (d x + c\right )^{2} - 128 \, a^{2} \cos \left (d x + c\right ) - 256 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*a^2*cos(d*x + c)^5 - 95*a^2*cos(d*x + c)^4 - 226*a^2*cos(d*x + c)^3 + 32*a^2*cos(d*x + c)^2 - 128*a^
2*cos(d*x + c) - 256*a^2 - (35*a^2*cos(d*x + c)^4 + 130*a^2*cos(d*x + c)^3 - 96*a^2*cos(d*x + c)^2 - 128*a^2*c
os(d*x + c) - 256*a^2)*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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giac [B]  time = 1.37, size = 306, normalized size = 2.41 \[ -\frac {1}{2520} \, \sqrt {2} {\left (\frac {252 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {1260 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {180 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {420 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {45 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} - \frac {420 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {35 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} - \frac {252 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} - \frac {3150 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \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*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/2520*sqrt(2)*(252*a^2*cos(1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 1260*a^2*cos(1/
4*pi + 1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 180*a^2*cos(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c))/d + 420*a^2*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
+ 45*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d - 420*a^2*sgn(cos(-1/4*pi + 1/2*d
*x + 1/2*c))*sin(1/4*pi + 3/2*d*x + 3/2*c)/d + 35*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*
x + 9/2*c)/d - 252*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 5/2*d*x + 5/2*c)/d - 3150*a^2*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.20, size = 77, normalized size = 0.61 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right )^{2} \left (35 \left (\sin ^{3}\left (d x +c \right )\right )+165 \left (\sin ^{2}\left (d x +c \right )\right )+321 \sin \left (d x +c \right )+319\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)^2*(35*sin(d*x+c)^3+165*sin(d*x+c)^2+321*sin(d*x+c)+319)/cos(d*x+c)/(a
+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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