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

Optimal. Leaf size=145 \[ -\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{6 a^2 d}+\frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}+\frac {13 \cos (c+d x)}{3 a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

1/2*cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(3/2)-11/4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+
c))^(1/2))/a^(3/2)/d*2^(1/2)+13/3*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-7/6*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/
a^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2765, 2968, 3023, 2751, 2649, 206} \[ -\frac {7 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{6 a^2 d}-\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}+\frac {13 \cos (c+d x)}{3 a d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + (Cos[c + d*x]
*Sin[c + d*x]^2)/(2*d*(a + a*Sin[c + d*x])^(3/2)) + (13*Cos[c + d*x])/(3*a*d*Sqrt[a + a*Sin[c + d*x]]) - (7*Co
s[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(6*a^2*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^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)]

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], 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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {\int \frac {\sin (c+d x) \left (2 a-\frac {7}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {\int \frac {2 a \sin (c+d x)-\frac {7}{2} a \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {7 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{6 a^2 d}-\frac {\int \frac {-\frac {7 a^2}{4}+\frac {13}{2} a^2 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{3 a^3}\\ &=\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac {13 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{6 a^2 d}+\frac {11 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{4 a}\\ &=\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac {13 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{6 a^2 d}-\frac {11 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{2 a d}\\ &=-\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac {13 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{6 a^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.26, size = 156, normalized size = 1.08 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (-11 \sin \left (\frac {1}{2} (c+d x)\right )+7 \sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {5}{2} (c+d x)\right )+11 \cos \left (\frac {1}{2} (c+d x)\right )+7 \cos \left (\frac {3}{2} (c+d x)\right )+\cos \left (\frac {5}{2} (c+d x)\right )+(33+33 i) (-1)^{3/4} (\sin (c+d x)+1) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )\right )}{6 d (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(11*Cos[(c + d*x)/2] + 7*Cos[(3*(c + d*x))/2] + Cos[(5*(c + d*x))/2] -
11*Sin[(c + d*x)/2] + (33 + 33*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(1 + Sin[
c + d*x]) + 7*Sin[(3*(c + d*x))/2] - Sin[(5*(c + d*x))/2]))/(6*d*(a*(1 + Sin[c + d*x]))^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.51, size = 295, normalized size = 2.03 \[ \frac {33 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (4 \, \cos \left (d x + c\right )^{3} + 16 \, \cos \left (d x + c\right )^{2} - {\left (4 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) + 15 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(33*sqrt(2)*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)*sqrt(a)*log(-(a*cos(d*x
+ c)^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a
*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)
) - 4*(4*cos(d*x + c)^3 + 16*cos(d*x + c)^2 - (4*cos(d*x + c)^2 - 12*cos(d*x + c) + 3)*sin(d*x + c) + 15*cos(d
*x + c) + 3)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d - (a^2*d*cos(d*x +
c) + 2*a^2*d)*sin(d*x + c))

________________________________________________________________________________________

giac [B]  time = 0.74, size = 409, normalized size = 2.82 \[ -\frac {\frac {8 \, {\left ({\left ({\left (\frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {3}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {2}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {33 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {6 \, {\left (3 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{3} + {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {a} - {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} a + a^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} \sqrt {a} - a\right )}^{2} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(8*(((2*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) - 3/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x
+ 1/2*c) + 3/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 2/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/
2*d*x + 1/2*c)^2 + a)^(3/2) - 33*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*
x + 1/2*c)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 6*(3*(sqrt(a)*tan(1/2*d*x
+ 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3 + (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a))^2*sqrt(a) - (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a + a^(3/2))/(((sqrt(
a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan
(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^2*a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

________________________________________________________________________________________

maple [A]  time = 0.68, size = 183, normalized size = 1.26 \[ -\frac {\left (\sin \left (d x +c \right ) \left (33 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-24 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right )+33 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-30 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{12 a^{\frac {7}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(sin(d*x+c)*(33*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2-8*(a-a*sin(d*x+c))^(3/2)
*a^(1/2)-24*(a-a*sin(d*x+c))^(1/2)*a^(3/2))+33*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2
-8*(a-a*sin(d*x+c))^(3/2)*a^(1/2)-30*(a-a*sin(d*x+c))^(1/2)*a^(3/2))*(-a*(sin(d*x+c)-1))^(1/2)/a^(7/2)/cos(d*x
+c)/(a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________