3.1.46 \(\int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [46]

Optimal. Leaf size=86 \[ -\frac {8 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d} \]

[Out]

-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-8/5*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/5*a*cos(d*x+c)*(a+a*sin
(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2830, 2726, 2725} \begin {gather*} -\frac {8 a^2 \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*Cos[c + d*x])/(5*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(5*d) - (2*
Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*d)

Rule 2725

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 2726

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] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2830

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*S
in[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))^{3/2} \, dx &=-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {3}{5} \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {1}{5} (4 a) \int \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {8 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 115, normalized size = 1.34 \begin {gather*} -\frac {(a (1+\sin (c+d x)))^{3/2} \left (20 \cos \left (\frac {1}{2} (c+d x)\right )+5 \cos \left (\frac {3}{2} (c+d x)\right )-\cos \left (\frac {5}{2} (c+d x)\right )-20 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )\right )}{10 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*((a*(1 + Sin[c + d*x]))^(3/2)*(20*Cos[(c + d*x)/2] + 5*Cos[(3*(c + d*x))/2] - Cos[(5*(c + d*x))/2] - 20*
Sin[(c + d*x)/2] + 5*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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Maple [A]
time = 1.84, size = 63, normalized size = 0.73

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (\sin ^{2}\left (d x +c \right )+3 \sin \left (d x +c \right )+6\right )}{5 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(sin(d*x+c)^2+3*sin(d*x+c)+6)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 99, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left (a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} - 7 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) - 4 \, a\right )} \sin \left (d x + c\right ) - 4 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 - 7*a*cos(d*x + c) - (a*cos(d*x + c)^2 + 3*a*cos(d*x + c) - 4*a)*si
n(d*x + c) - 4*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**(3/2)*sin(c + d*x), x)

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Giac [A]
time = 0.67, size = 95, normalized size = 1.10 \begin {gather*} \frac {\sqrt {2} {\left (20 \, a \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 ) + 5 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )} \sqrt {a}}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*sqrt(2)*(20*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 5*a*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d*x
+ 5/2*c))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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