\(\int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [44]

Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {68 a^2 \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {34 a^2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {136 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {68 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 d} \]

[Out]

-68/105*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-68/45*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-34/63*a^2*cos(d*x+c)
*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/9*a^2*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+136/315*a*cos(
d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2842, 21, 2849, 2838, 2830, 2725} \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}-\frac {34 a^2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {68 a^2 \cos (c+d x)}{45 d \sqrt {a \sin (c+d x)+a}}-\frac {68 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 d}+\frac {136 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 d} \]

[In]

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

[Out]

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

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 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 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)]

Rule 2838

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 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 2849

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]

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

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-3780-4830 \cos (c+d x)+4352 \sqrt {2} \sqrt {1+\cos (c+d x)}-672 \cos (2 (c+d x))+513 \cos (3 (c+d x))+100 \cos (4 (c+d x))-35 \cos (5 (c+d x))+2730 \sin (c+d x)-1428 \sin (2 (c+d x))-243 \sin (3 (c+d x))+170 \sin (4 (c+d x))+35 \sin (5 (c+d x))\right )}{5040 d \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(a*Sec[(c + d*x)/2]^2*Sqrt[a*(1 + Sin[c + d*x])]*(-3780 - 4830*Cos[c + d*x] + 4352*Sqrt[2]*Sqrt[1 + Cos[c + d*
x]] - 672*Cos[2*(c + d*x)] + 513*Cos[3*(c + d*x)] + 100*Cos[4*(c + d*x)] - 35*Cos[5*(c + d*x)] + 2730*Sin[c +
d*x] - 1428*Sin[2*(c + d*x)] - 243*Sin[3*(c + d*x)] + 170*Sin[4*(c + d*x)] + 35*Sin[5*(c + d*x)]))/(5040*d*(1
+ Tan[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (35 \left (\sin ^{4}\left (d x +c \right )\right )+85 \left (\sin ^{3}\left (d x +c \right )\right )+102 \left (\sin ^{2}\left (d x +c \right )\right )+136 \sin \left (d x +c \right )+272\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(85\)

[In]

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

[Out]

2/315*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+85*sin(d*x+c)^3+102*sin(d*x+c)^2+136*sin(d*x+c)+272)/
cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.90 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 172 \, a \cos \left (d x + c\right )^{3} + 134 \, a \cos \left (d x + c\right )^{2} + 409 \, a \cos \left (d x + c\right ) - {\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} - 87 \, a \cos \left (d x + c\right )^{2} - 221 \, a \cos \left (d x + c\right ) + 188 \, a\right )} \sin \left (d x + c\right ) + 188 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

-2/315*(35*a*cos(d*x + c)^5 - 50*a*cos(d*x + c)^4 - 172*a*cos(d*x + c)^3 + 134*a*cos(d*x + c)^2 + 409*a*cos(d*
x + c) - (35*a*cos(d*x + c)^4 + 85*a*cos(d*x + c)^3 - 87*a*cos(d*x + c)^2 - 221*a*cos(d*x + c) + 188*a)*sin(d*
x + c) + 188*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

Sympy [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(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)^3, x)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (3780 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1050 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 378 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 135 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

[In]

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

[Out]

1/2520*sqrt(2)*(3780*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1050*a*sgn(cos(-1/
4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 378*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*p
i + 5/2*d*x + 5/2*c) + 135*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) + 35*a*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 9/2*c))*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

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

[Out]

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