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

Optimal. Leaf size=91 \[ \frac{a \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{b \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.102029, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3379, 2664, 12, 2660, 618, 204} \[ \frac{a \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{b \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sin[c + d*x^2])^2,x]

[Out]

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

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]))

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 12

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.205112, size = 91, normalized size = 1. \[ \frac{\frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{b \cos \left (c+d x^2\right )}{a+b \sin \left (c+d x^2\right )}}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.041, size = 164, normalized size = 1.8 \begin{align*}{\frac{{b}^{2}}{da \left ({a}^{2}-{b}^{2} \right ) }\tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) b+a \right ) ^{-1}}+{\frac{b}{d \left ({a}^{2}-{b}^{2} \right ) } \left ( \left ( \tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) b+a \right ) ^{-1}}+{\frac{a}{d}\arctan \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) +2\,b \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 2.10745, size = 798, normalized size = 8.77 \begin{align*} \left [\frac{{\left (a b \sin \left (d x^{2} + c\right ) + a^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (d x^{2} + c\right ) \sin \left (d x^{2} + c\right ) + b \cos \left (d x^{2} + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x^{2} + c\right )}{4 \,{\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{2} + c\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, -\frac{{\left (a b \sin \left (d x^{2} + c\right ) + a^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x^{2} + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x^{2} + c\right )}\right ) -{\left (a^{2} b - b^{3}\right )} \cos \left (d x^{2} + c\right )}{2 \,{\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{2} + c\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.11736, size = 194, normalized size = 2.13 \begin{align*} \frac{{\left (\pi \left \lfloor \frac{d x^{2} + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a}{{\left (a^{2} d - b^{2} d\right )} \sqrt{a^{2} - b^{2}}} + \frac{b^{2} \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + a b}{{\left (a^{3} d - a b^{2} d\right )}{\left (a \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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