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

Optimal. Leaf size=94 \[ \frac {2 a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} \left (c+d x^3\right )\right )+b}{\sqrt {a^2-b^2}}\right )}{3 d \left (a^2-b^2\right )^{3/2}}+\frac {b \cos \left (c+d x^3\right )}{3 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )} \]

[Out]

2/3*a*arctan((b+a*tan(1/2*d*x^3+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)/d+1/3*b*cos(d*x^3+c)/(a^2-b^2)/d/(a+b
*sin(d*x^3+c))

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3379, 2664, 12, 2660, 618, 204} \[ \frac {2 a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} \left (c+d x^3\right )\right )+b}{\sqrt {a^2-b^2}}\right )}{3 d \left (a^2-b^2\right )^{3/2}}+\frac {b \cos \left (c+d x^3\right )}{3 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*Sin[c + d*x^3])^2,x]

[Out]

(2*a*ArcTan[(b + a*Tan[(c + d*x^3)/2])/Sqrt[a^2 - b^2]])/(3*(a^2 - b^2)^(3/2)*d) + (b*Cos[c + d*x^3])/(3*(a^2
- b^2)*d*(a + b*Sin[c + d*x^3]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2664

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 + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 91, normalized size = 0.97 \[ \frac {\frac {2 a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} \left (c+d x^3\right )\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {b \cos \left (c+d x^3\right )}{a+b \sin \left (c+d x^3\right )}}{3 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*ArcTan[(b + a*Tan[(c + d*x^3)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (b*Cos[c + d*x^3])/(a + b*Sin[c +
d*x^3]))/(3*(a - b)*(a + b)*d)

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 366, normalized size = 3.89 \[ \left [\frac {{\left (a b \sin \left (d x^{3} + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x^{3} + c\right )^{2} - 2 \, a b \sin \left (d x^{3} + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x^{3} + c\right ) \sin \left (d x^{3} + c\right ) + b \cos \left (d x^{3} + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, a b \sin \left (d x^{3} + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x^{3} + c\right )}{6 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{3} + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, -\frac {{\left (a b \sin \left (d x^{3} + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x^{3} + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x^{3} + c\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (d x^{3} + c\right )}{3 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{3} + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*((a*b*sin(d*x^3 + c) + a^2)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x^3 + c)^2 - 2*a*b*sin(d*x^3 + c)
- a^2 - b^2 - 2*(a*cos(d*x^3 + c)*sin(d*x^3 + c) + b*cos(d*x^3 + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x^3 + c)^2 -
2*a*b*sin(d*x^3 + c) - a^2 - b^2)) + 2*(a^2*b - b^3)*cos(d*x^3 + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*sin(d*x^3 +
c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -1/3*((a*b*sin(d*x^3 + c) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x^3 + c)
+ b)/(sqrt(a^2 - b^2)*cos(d*x^3 + c))) - (a^2*b - b^3)*cos(d*x^3 + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*sin(d*x^3
+ c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)]

________________________________________________________________________________________

giac [A]  time = 1.89, size = 146, normalized size = 1.55 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {d x^{3} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{3 \, {\left (a^{2} d - b^{2} d\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right ) + a b\right )}}{3 \, {\left (a^{3} d - a b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right ) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(pi*floor(1/2*(d*x^3 + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x^3 + 1/2*c) + b)/sqrt(a^2 - b^2)))*a/((a
^2*d - b^2*d)*sqrt(a^2 - b^2)) + 2/3*(b^2*tan(1/2*d*x^3 + 1/2*c) + a*b)/((a^3*d - a*b^2*d)*(a*tan(1/2*d*x^3 +
1/2*c)^2 + 2*b*tan(1/2*d*x^3 + 1/2*c) + a))

________________________________________________________________________________________

maple [A]  time = 0.10, size = 167, normalized size = 1.78 \[ \frac {2 b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d \left (a \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right ) b +a \right ) a \left (a^{2}-b^{2}\right )}+\frac {2 b}{3 d \left (a \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{3 d \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d/(a*tan(1/2*d*x^3+1/2*c)^2+2*tan(1/2*d*x^3+1/2*c)*b+a)*b^2/a/(a^2-b^2)*tan(1/2*d*x^3+1/2*c)+2/3/d/(a*tan(
1/2*d*x^3+1/2*c)^2+2*tan(1/2*d*x^3+1/2*c)*b+a)*b/(a^2-b^2)+2/3/d*a/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x
^3+1/2*c)+2*b)/(a^2-b^2)^(1/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 4.95, size = 186, normalized size = 1.98 \[ \frac {\frac {2\,b}{a^2-b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )}{a\,\left (a^2-b^2\right )}}{d\,\left (3\,a\,{\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )}^2+6\,b\,\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )+3\,a\right )}+\frac {2\,a\,\mathrm {atan}\left (\frac {3\,\left (a^2-b^2\right )\,\left (\frac {2\,a^2\,\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )}{3\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\left (3\,a^2\,b-3\,b^3\right )}{9\,{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{2\,a}\right )}{3\,d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*sin(c + d*x^3))^2,x)

[Out]

((2*b)/(a^2 - b^2) + (2*b^2*tan(c/2 + (d*x^3)/2))/(a*(a^2 - b^2)))/(d*(3*a + 3*a*tan(c/2 + (d*x^3)/2)^2 + 6*b*
tan(c/2 + (d*x^3)/2))) + (2*a*atan((3*(a^2 - b^2)*((2*a^2*tan(c/2 + (d*x^3)/2))/(3*(a + b)^(3/2)*(a - b)^(3/2)
) + (2*a*(3*a^2*b - 3*b^3))/(9*(a + b)^(3/2)*(a^2 - b^2)*(a - b)^(3/2))))/(2*a)))/(3*d*(a + b)^(3/2)*(a - b)^(
3/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________