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

Optimal result

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

-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-68/105*cos(d* 
x+c)*(a+a*sin(d*x+c))^(3/2)/d
 

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-3780-4830 \cos (c+d x)+4352 \sqrt {2} \sqrt {1+\cos (c+d x)}-672 \cos (2 (c+d x))+513 \cos (3 (c+d x))+100 \cos (4 (c+d x))-35 \cos (5 (c+d x))+2730 \sin (c+d x)-1428 \sin (2 (c+d x))-243 \sin (3 (c+d x))+170 \sin (4 (c+d x))+35 \sin (5 (c+d x))\right )}{5040 d \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

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

Output:

(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]))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3242, 27, 2011, 3042, 3249, 3042, 3238, 27, 3042, 3230, 3042, 3125}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^3 (a \sin (c+d x)+a)^{3/2}dx\)

\(\Big \downarrow \) 3242

\(\displaystyle \frac {2}{9} \int \frac {17 \sin ^3(c+d x) \left (\sin (c+d x) a^2+a^2\right )}{2 \sqrt {\sin (c+d x) a+a}}dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {17}{9} \int \frac {\sin ^3(c+d x) \left (\sin (c+d x) a^2+a^2\right )}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {17}{9} a \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {17}{9} a \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \int \sin ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \int \sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3238

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {17}{9} a \left (\frac {6}{7} \left (\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

Input:

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

Output:

(-2*a^2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (17* 
a*((-2*a*Cos[c + d*x]*Sin[c + d*x]^3)/(7*d*Sqrt[a + a*Sin[c + d*x]]) + (6* 
((-2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*d) + ((-14*a^2*Cos[c + 
d*x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (4*a*Cos[c + d*x]*Sqrt[a + a*Sin[c 
+ d*x]])/(3*d))/(5*a)))/7))/9
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3125

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] && Eq 
Q[a^2 - b^2, 0]
 

rule 3230

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] + Simp[(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 3238

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] + Simp[1/(b*(m + 2))   Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*Si 
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  ! 
LtQ[m, -2^(-1)]
 

rule 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] + Simp[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 3249

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] + Simp[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]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.52

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

Input:

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

Output:

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+1 
02*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)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.90 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 172 \, a \cos \left (d x + c\right )^{3} + 134 \, a \cos \left (d x + c\right )^{2} + 409 \, a \cos \left (d x + c\right ) - {\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} - 87 \, a \cos \left (d x + c\right )^{2} - 221 \, a \cos \left (d x + c\right ) + 188 \, a\right )} \sin \left (d x + c\right ) + 188 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

-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*c 
os(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (3780 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1050 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 378 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 135 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \] Input:

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

Output:

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

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

Output:

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

Reduce [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{4}d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{3}d x \right ) \] Input:

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

Output:

sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*sin(c + d*x)**4,x) + int(sqrt(sin(c 
+ d*x) + 1)*sin(c + d*x)**3,x))