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

Optimal. Leaf size=116 \[ -\frac {64 a^3 \cos (c+d x)}{21 d \sqrt {a \sin (c+d x)+a}}-\frac {16 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{21 d}-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{7 d}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d} \]

[Out]

-2/7*a*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/7*cos(d*x+c)*(a+a*sin(d*x+c))^(5/2)/d-64/21*a^3*cos(d*x+c)/d/(a+a
*sin(d*x+c))^(1/2)-16/21*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2751, 2647, 2646} \[ -\frac {64 a^3 \cos (c+d x)}{21 d \sqrt {a \sin (c+d x)+a}}-\frac {16 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{21 d}-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{7 d}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x])/(21*d*Sqrt[a + a*Sin[c + d*x]]) - (16*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(21*d)
- (2*a*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(7*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*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)]

Rubi steps

\begin {align*} \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}+\frac {5}{7} \int (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac {2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{7 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}+\frac {1}{7} (8 a) \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {16 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{7 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}+\frac {1}{21} \left (32 a^2\right ) \int \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {64 a^3 \cos (c+d x)}{21 d \sqrt {a+a \sin (c+d x)}}-\frac {16 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{7 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 141, normalized size = 1.22 \[ \frac {(a (\sin (c+d x)+1))^{5/2} \left (315 \sin \left (\frac {1}{2} (c+d x)\right )-77 \sin \left (\frac {3}{2} (c+d x)\right )-21 \sin \left (\frac {5}{2} (c+d x)\right )+3 \sin \left (\frac {7}{2} (c+d x)\right )-315 \cos \left (\frac {1}{2} (c+d x)\right )-77 \cos \left (\frac {3}{2} (c+d x)\right )+21 \cos \left (\frac {5}{2} (c+d x)\right )+3 \cos \left (\frac {7}{2} (c+d x)\right )\right )}{84 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(5/2)*(-315*Cos[(c + d*x)/2] - 77*Cos[(3*(c + d*x))/2] + 21*Cos[(5*(c + d*x))/2] + 3*C
os[(7*(c + d*x))/2] + 315*Sin[(c + d*x)/2] - 77*Sin[(3*(c + d*x))/2] - 21*Sin[(5*(c + d*x))/2] + 3*Sin[(7*(c +
d*x))/2]))/(84*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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fricas [A]  time = 0.47, size = 140, normalized size = 1.21 \[ \frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} + 12 \, a^{2} \cos \left (d x + c\right )^{3} - 17 \, a^{2} \cos \left (d x + c\right )^{2} - 58 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2} + {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 9 \, a^{2} \cos \left (d x + c\right )^{2} - 26 \, a^{2} \cos \left (d x + c\right ) + 32 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{21 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*(3*a^2*cos(d*x + c)^4 + 12*a^2*cos(d*x + c)^3 - 17*a^2*cos(d*x + c)^2 - 58*a^2*cos(d*x + c) - 32*a^2 + (3
*a^2*cos(d*x + c)^3 - 9*a^2*cos(d*x + c)^2 - 26*a^2*cos(d*x + c) + 32*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 = 0.73, size = 240, normalized size = 2.07 \[ \frac {1}{420} \, \sqrt {2} {\left (\frac {21 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {735 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {15 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {245 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {140 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} - \frac {84 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {840 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*sqrt(2)*(21*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 - 735*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 + 15*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 - 245*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 - 140
*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 - 84*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 + 840*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 = 0.61, size = 75, normalized size = 0.65 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (3 \left (\sin ^{3}\left (d x +c \right )\right )+12 \left (\sin ^{2}\left (d x +c \right )\right )+23 \sin \left (d x +c \right )+46\right )}{21 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(3*sin(d*x+c)^3+12*sin(d*x+c)^2+23*sin(d*x+c)+46)/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sin \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \sin {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**(5/2)*sin(c + d*x), x)

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