3.1.37 \(\int \frac {x}{a+b \sin (c+d x^2)} \, dx\) [37]

Optimal. Leaf size=48 \[ \frac {\tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3460, 2739, 632, 210} \begin {gather*} \frac {\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 \sqrt {a^2-b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sin[c + d*x^2]),x]

[Out]

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

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

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Mathematica [A]
time = 0.06, size = 48, normalized size = 1.00 \begin {gather*} \frac {\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} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Sin[c + d*x^2]),x]

[Out]

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

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Maple [A]
time = 0.06, size = 48, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}}\)	\(48\)
default	\(\frac {\arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}}\)	\(48\)
risch	\(-\frac {\ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/(a^2-b^2)^(1/2)*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 [B] Leaf count of result is larger than twice the leaf count of optimal. 8078 vs. \(2 (43) = 86\).
time = 28.31, size = 8078, normalized size = 168.29 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan2(-2*(4*(a^2*b^4 - b^6)*cos(d*x^2 + 2*c)^4*cos(c)*sin(c) - 4*(a^2*b^4 - b^6)*cos(c)*sin(d*x^2 + 2*c)
^4*sin(c) - 4*((a^3*b^3 - a*b^5)*cos(c)^3 + 3*(a^3*b^3 - a*b^5)*cos(c)*sin(c)^2)*cos(d*x^2 + 2*c)^3 - 4*(3*(a^
3*b^3 - a*b^5)*cos(c)^2*sin(c) + (a^3*b^3 - a*b^5)*sin(c)^3 + ((a^2*b^4 - b^6)*cos(c)^2 - (a^2*b^4 - b^6)*sin(
c)^2)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c)^3 + 4*((4*a^4*b^2 - 5*a^2*b^4 + b^6)*cos(c)^3*sin(c) + (4*a^4*b^2 - 5
*a^2*b^4 + b^6)*cos(c)*sin(c)^3)*cos(d*x^2 + 2*c)^2 - 4*((4*a^4*b^2 - 5*a^2*b^4 + b^6)*cos(c)^3*sin(c) + (4*a^
4*b^2 - 5*a^2*b^4 + b^6)*cos(c)*sin(c)^3 + 3*((a^3*b^3 - a*b^5)*cos(c)^3 - (a^3*b^3 - a*b^5)*cos(c)*sin(c)^2)*
cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c)^2 - 4*((2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(c)^5 + 2*(2*a^5*b - 3*a^3*b^3 + a*
b^5)*cos(c)^3*sin(c)^2 + (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(c)*sin(c)^4)*cos(d*x^2 + 2*c) - 4*((2*a^5*b - 3*a^3
*b^3 + a*b^5)*cos(c)^4*sin(c) + 2*(2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(c)^2*sin(c)^3 + (2*a^5*b - 3*a^3*b^3 + a*b
^5)*sin(c)^5 + ((a^2*b^4 - b^6)*cos(c)^2 - (a^2*b^4 - b^6)*sin(c)^2)*cos(d*x^2 + 2*c)^3 - 3*((a^3*b^3 - a*b^5)
*cos(c)^2*sin(c) - (a^3*b^3 - a*b^5)*sin(c)^3)*cos(d*x^2 + 2*c)^2 + ((4*a^4*b^2 - 5*a^2*b^4 + b^6)*cos(c)^4 -
(4*a^4*b^2 - 5*a^2*b^4 + b^6)*sin(c)^4)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c) + (b^5*cos(d*x^2 + 2*c)^5*cos(c) -
4*a*b^4*cos(d*x^2 + 2*c)^4*cos(c)*sin(c) + b^5*sin(d*x^2 + 2*c)^5*sin(c) + (b^5*cos(d*x^2 + 2*c)*cos(c) + 4*a*
b^4*cos(c)*sin(c))*sin(d*x^2 + 2*c)^4 + 2*((2*a^2*b^3 - b^5)*cos(c)^3 + 3*(2*a^2*b^3 - b^5)*cos(c)*sin(c)^2)*c
os(d*x^2 + 2*c)^3 + 2*(b^5*cos(d*x^2 + 2*c)^2*sin(c) + 3*(2*a^2*b^3 - b^5)*cos(c)^2*sin(c) + (2*a^2*b^3 - b^5)
*sin(c)^3 + 2*(a*b^4*cos(c)^2 - a*b^4*sin(c)^2)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c)^3 - 4*((4*a^3*b^2 - 3*a*b^4
)*cos(c)^3*sin(c) + (4*a^3*b^2 - 3*a*b^4)*cos(c)*sin(c)^3)*cos(d*x^2 + 2*c)^2 + 2*(b^5*cos(d*x^2 + 2*c)^3*cos(
c) + 2*(4*a^3*b^2 - 3*a*b^4)*cos(c)^3*sin(c) + 2*(4*a^3*b^2 - 3*a*b^4)*cos(c)*sin(c)^3 + 3*((2*a^2*b^3 - b^5)*
cos(c)^3 - (2*a^2*b^3 - b^5)*cos(c)*sin(c)^2)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c)^2 + ((8*a^4*b - 8*a^2*b^3 + b
^5)*cos(c)^5 + 2*(8*a^4*b - 8*a^2*b^3 + b^5)*cos(c)^3*sin(c)^2 + (8*a^4*b - 8*a^2*b^3 + b^5)*cos(c)*sin(c)^4)*
cos(d*x^2 + 2*c) + (b^5*cos(d*x^2 + 2*c)^4*sin(c) + (8*a^4*b - 8*a^2*b^3 + b^5)*cos(c)^4*sin(c) + 2*(8*a^4*b -
8*a^2*b^3 + b^5)*cos(c)^2*sin(c)^3 + (8*a^4*b - 8*a^2*b^3 + b^5)*sin(c)^5 + 4*(a*b^4*cos(c)^2 - a*b^4*sin(c)^
2)*cos(d*x^2 + 2*c)^3 - 6*((2*a^2*b^3 - b^5)*cos(c)^2*sin(c) - (2*a^2*b^3 - b^5)*sin(c)^3)*cos(d*x^2 + 2*c)^2
+ 4*((4*a^3*b^2 - 3*a*b^4)*cos(c)^4 - (4*a^3*b^2 - 3*a*b^4)*sin(c)^4)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c))*sqrt
(a^2 - b^2))/(b^6*cos(d*x^2 + 2*c)^6 + 6*a*b^5*cos(c)*sin(d*x^2 + 2*c)^5 + b^6*sin(d*x^2 + 2*c)^6 - 6*a*b^5*co
s(d*x^2 + 2*c)^5*sin(c) + (32*a^6 - 48*a^4*b^2 + 18*a^2*b^4 - b^6)*cos(c)^6 + 3*(32*a^6 - 48*a^4*b^2 + 18*a^2*
b^4 - b^6)*cos(c)^4*sin(c)^2 + 3*(32*a^6 - 48*a^4*b^2 + 18*a^2*b^4 - b^6)*cos(c)^2*sin(c)^4 + (32*a^6 - 48*a^4
*b^2 + 18*a^2*b^4 - b^6)*sin(c)^6 + 3*((2*a^2*b^4 - b^6)*cos(c)^2 + 5*(2*a^2*b^4 - b^6)*sin(c)^2)*cos(d*x^2 +
2*c)^4 + 3*(b^6*cos(d*x^2 + 2*c)^2 - 2*a*b^5*cos(d*x^2 + 2*c)*sin(c) + 5*(2*a^2*b^4 - b^6)*cos(c)^2 + (2*a^2*b
^4 - b^6)*sin(c)^2)*sin(d*x^2 + 2*c)^4 - 4*(3*(4*a^3*b^3 - 3*a*b^5)*cos(c)^2*sin(c) + 5*(4*a^3*b^3 - 3*a*b^5)*
sin(c)^3)*cos(d*x^2 + 2*c)^3 + 4*(3*a*b^5*cos(d*x^2 + 2*c)^2*cos(c) + 5*(4*a^3*b^3 - 3*a*b^5)*cos(c)^3 - 6*(2*
a^2*b^4 - b^6)*cos(d*x^2 + 2*c)*cos(c)*sin(c) + 3*(4*a^3*b^3 - 3*a*b^5)*cos(c)*sin(c)^2)*sin(d*x^2 + 2*c)^3 +
3*((8*a^4*b^2 - 8*a^2*b^4 + b^6)*cos(c)^4 + 6*(8*a^4*b^2 - 8*a^2*b^4 + b^6)*cos(c)^2*sin(c)^2 + 5*(8*a^4*b^2 -
8*a^2*b^4 + b^6)*sin(c)^4)*cos(d*x^2 + 2*c)^2 + 3*(b^6*cos(d*x^2 + 2*c)^4 - 4*a*b^5*cos(d*x^2 + 2*c)^3*sin(c)
+ 5*(8*a^4*b^2 - 8*a^2*b^4 + b^6)*cos(c)^4 + 6*(8*a^4*b^2 - 8*a^2*b^4 + b^6)*cos(c)^2*sin(c)^2 + (8*a^4*b^2 -
8*a^2*b^4 + b^6)*sin(c)^4 + 6*((2*a^2*b^4 - b^6)*cos(c)^2 + (2*a^2*b^4 - b^6)*sin(c)^2)*cos(d*x^2 + 2*c)^2 -
4*(3*(4*a^3*b^3 - 3*a*b^5)*cos(c)^2*sin(c) + (4*a^3*b^3 - 3*a*b^5)*sin(c)^3)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c
)^2 - 6*((16*a^5*b - 20*a^3*b^3 + 5*a*b^5)*cos(c)^4*sin(c) + 2*(16*a^5*b - 20*a^3*b^3 + 5*a*b^5)*cos(c)^2*sin(
c)^3 + (16*a^5*b - 20*a^3*b^3 + 5*a*b^5)*sin(c)^5)*cos(d*x^2 + 2*c) + 6*(a*b^5*cos(d*x^2 + 2*c)^4*cos(c) + (16
*a^5*b - 20*a^3*b^3 + 5*a*b^5)*cos(c)^5 - 4*(2*a^2*b^4 - b^6)*cos(d*x^2 + 2*c)^3*cos(c)*sin(c) + 2*(16*a^5*b -
20*a^3*b^3 + 5*a*b^5)*cos(c)^3*sin(c)^2 + (16*a^5*b - 20*a^3*b^3 + 5*a*b^5)*cos(c)*sin(c)^4 + 2*((4*a^3*b^3 -
3*a*b^5)*cos(c)^3 + 3*(4*a^3*b^3 - 3*a*b^5)*cos(c)*sin(c)^2)*cos(d*x^2 + 2*c)^2 - 4*((8*a^4*b^2 - 8*a^2*b^4 +
b^6)*cos(c)^3*sin(c) + (8*a^4*b^2 - 8*a^2*b^4 + b^6)*cos(c)*sin(c)^3)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c) - 2*
(3*b^5*cos(c)*sin(d*x^2 + 2*c)^5 - 3*b^5*cos(d*x^2 + 2*c)^5*sin(c) + (16*a^5 - 16*a^3*b^2 + 3*a*b^4)*cos(c)^6
+ 3*(16*a^5 - 16*a^3*b^2 + 3*a*b^4)*cos(c)^4*sin(c)^2 + 3*(16*a^5 - 16*a^3*b^2 + 3*a*b^4)*cos(c)^2*sin(c)^4 +
(16*a^5 - 16*a^3*b^2 + 3*a*b^4)*sin(c)^6 + 3*(a...

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Fricas [A]
time = 0.36, size = 208, normalized size = 4.33 \begin {gather*} \left [-\frac {\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 )}{4 \, {\left (a^{2} - b^{2}\right )} d}, -\frac {\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 )}{2 \, \sqrt {a^{2} - b^{2}} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (37) = 74\).
time = 5.72, size = 202, normalized size = 4.21 \begin {gather*} \begin {cases} \frac {\tilde {\infty } x^{2}}{\sin {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} \right )}}{2 b d} & \text {for}\: a = 0 \\\frac {x^{2}}{2 \left (a + b \sin {\left (c \right )}\right )} & \text {for}\: d = 0 \\\frac {\sqrt {b^{2}}}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} - b d \sqrt {b^{2}}} & \text {for}\: a = - \sqrt {b^{2}} \\- \frac {\sqrt {b^{2}}}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} + b d \sqrt {b^{2}}} & \text {for}\: a = \sqrt {b^{2}} \\\frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{2 d \sqrt {- a^{2} + b^{2}}} - \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{2 d \sqrt {- a^{2} + b^{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**2/sin(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (log(tan(c/2 + d*x**2/2))/(2*b*d), Eq(a, 0)), (x*
*2/(2*(a + b*sin(c))), Eq(d, 0)), (sqrt(b**2)/(b**2*d*tan(c/2 + d*x**2/2) - b*d*sqrt(b**2)), Eq(a, -sqrt(b**2)
)), (-sqrt(b**2)/(b**2*d*tan(c/2 + d*x**2/2) + b*d*sqrt(b**2)), Eq(a, sqrt(b**2))), (log(tan(c/2 + d*x**2/2) +
b/a - sqrt(-a**2 + b**2)/a)/(2*d*sqrt(-a**2 + b**2)) - log(tan(c/2 + d*x**2/2) + b/a + sqrt(-a**2 + b**2)/a)/
(2*d*sqrt(-a**2 + b**2)), True))

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Giac [A]
time = 5.44, size = 63, normalized size = 1.31 \begin {gather*} \frac {\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 )}{\sqrt {a^{2} - b^{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 6.63, size = 128, normalized size = 2.67 \begin {gather*} -\frac {\ln \left (-x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}-\frac {2\,x\,\left (b\,1{}\mathrm {i}+a\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {b-a}}\right )-\ln \left (-x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}+\frac {2\,x\,\left (b\,1{}\mathrm {i}+a\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {b-a}}\right )}{2\,d\,\sqrt {a+b}\,\sqrt {b-a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*sin(c + d*x^2)),x)

[Out]

-(log(- x*exp(d*x^2*1i)*exp(c*1i)*2i - (2*x*(b*1i + a*exp(d*x^2*1i)*exp(c*1i)))/((a + b)^(1/2)*(b - a)^(1/2)))
- log((2*x*(b*1i + a*exp(d*x^2*1i)*exp(c*1i)))/((a + b)^(1/2)*(b - a)^(1/2)) - x*exp(d*x^2*1i)*exp(c*1i)*2i))
/(2*d*(a + b)^(1/2)*(b - a)^(1/2))

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