3.1.45 \(\int \frac {x}{(a+b \sin (c+d x^2))^2} \, dx\) [45]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 = {3460, 2743, 12, 2739, 632, 210} \begin {gather*} \frac {a \text {ArcTan}\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 )} \end {gather*}

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 12

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

Rule 210

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

Rule 632

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 2739

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 2743

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] && Integ
erQ[2*n]

Rule 3460

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}{\left (a+b \sin \left (c+d x^2\right )\right )^2} \, dx &=\frac {1}{2} \text {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 {\text {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 \text {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 \text {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) \text {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*}

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Mathematica [A]
time = 0.15, size = 91, normalized size = 1.00 \begin {gather*} \frac {\frac {2 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 )}{\sqrt {a^2-b^2}}+\frac {b \cos \left (c+d x^2\right )}{a+b \sin \left (c+d x^2\right )}}{2 (a-b) (a+b) d} \end {gather*}

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)

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Maple [A]
time = 0.14, size = 131, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{a \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{2 d}\)	\(131\)
default	\(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{a \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{2 d}\)	\(131\)
risch	\(\frac {i b +a \,{\mathrm e}^{i \left (d \,x^{2}+c \right )}}{\left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d \,x^{2}+c \right )}-b +2 i a \,{\mathrm e}^{i \left (d \,x^{2}+c \right )}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Fricas [A]
time = 0.41, size = 366, normalized size = 4.02 \begin {gather*} \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 {gather*}

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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.35, size = 144, normalized size = 1.58 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 5.10, size = 178, normalized size = 1.96 \begin {gather*} \frac {\frac {b}{a^2-b^2}+\frac {b^2\,\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}{a\,\left (a^2-b^2\right )}}{d\,\left (a\,{\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )+a\right )}+\frac {a\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {a^2\,\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {a\,\left (2\,a^2\,b-2\,b^3\right )}{2\,{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{a}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*sin(c + d*x^2))^2,x)

[Out]

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

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