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

Optimal. Leaf size=245 \[ -\frac {\text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d^2 \sqrt {a^2-b^2}}+\frac {\text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 d^2 \sqrt {a^2-b^2}}-\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d \sqrt {a^2-b^2}}+\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d \sqrt {a^2-b^2}} \]

[Out]

-1/2*I*x^2*ln(1-I*b*exp(I*(d*x^2+c))/(a-(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)+1/2*I*x^2*ln(1-I*b*exp(I*(d*x^2+c)
)/(a+(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)-1/2*polylog(2,I*b*exp(I*(d*x^2+c))/(a-(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)
^(1/2)+1/2*polylog(2,I*b*exp(I*(d*x^2+c))/(a+(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.51, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3379, 3323, 2264, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d^2 \sqrt {a^2-b^2}}+\frac {\text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d^2 \sqrt {a^2-b^2}}-\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d \sqrt {a^2-b^2}}+\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*Sin[c + d*x^2]),x]

[Out]

((-I/2)*x^2*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + ((I/2)*x^2*Log[1 - (
I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a - S
qrt[a^2 - b^2])]/(2*Sqrt[a^2 - b^2]*d^2) + PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])]/(2*Sqrt[a
^2 - b^2]*d^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3323

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

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^3}{a+b \sin \left (c+d x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{a+b \sin (c+d x)} \, dx,x,x^2\right )\\ &=\operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx,x,x^2\right )\\ &=-\frac {(i b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {a^2-b^2}}+\frac {(i b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {a^2-b^2}}\\ &=-\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}+\frac {i \operatorname {Subst}\left (\int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2-b^2} d}-\frac {i \operatorname {Subst}\left (\int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2-b^2} d}\\ &=-\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 \sqrt {a^2-b^2} d^2}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 \sqrt {a^2-b^2} d^2}\\ &=-\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}-\frac {\text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d^2}+\frac {\text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 188, normalized size = 0.77 \[ \frac {-\text {Li}_2\left (-\frac {i b e^{i \left (d x^2+c\right )}}{\sqrt {a^2-b^2}-a}\right )+\text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )-i d x^2 \left (\log \left (1+\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}-a}\right )-\log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )\right )}{2 d^2 \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*d*x^2*(Log[1 + (I*b*E^(I*(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])] - Log[1 - (I*b*E^(I*(c + d*x^2)))/(a + Sq
rt[a^2 - b^2])]) - PolyLog[2, ((-I)*b*E^(I*(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])] + PolyLog[2, (I*b*E^(I*(c + d
*x^2)))/(a + Sqrt[a^2 - b^2])])/(2*Sqrt[a^2 - b^2]*d^2)

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fricas [B]  time = 0.98, size = 1053, normalized size = 4.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*b*c*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2)
+ 2*I*a) + 2*b*c*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/
b^2) - 2*I*a) - 2*b*c*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 -
b^2)/b^2) + 2*I*a) - 2*b*c*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-
(a^2 - b^2)/b^2) - 2*I*a) - 2*I*b*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c)
+ 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*b*sqrt(-(a^2 - b^2)/b^
2)*dilog(-1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^
2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*b*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 +
c) + 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*I*b*sqrt(-(a^2 - b^2)
/b^2)*dilog(-1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(
-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*(b*d*x^2 + b*c)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x^2 + c) + 2*a
*sin(d*x^2 + c) + 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(b*d*x^2 + b*
c)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) - I*b*sin(d
*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b*d*x^2 + b*c)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x
^2 + c) + 2*a*sin(d*x^2 + c) + 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*
(b*d*x^2 + b*c)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 +
c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b))/((a^2 - b^2)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{a +b \sin \left (d \,x^{2}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*sin(c + d*x^2)),x)

[Out]

int(x^3/(a + b*sin(c + d*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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