3.1.81 \(\int \frac {x^5}{a+b \sin (c+d x^3)} \, dx\) [81]

Optimal. Leaf size=245 \[ -\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}+\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}-\frac {\text {Li}_2\left (\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d^2}+\frac {\text {Li}_2\left (\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d^2} \]

[Out]

-1/3*I*x^3*ln(1-I*b*exp(I*(d*x^3+c))/(a-(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)+1/3*I*x^3*ln(1-I*b*exp(I*(d*x^3+c)
)/(a+(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)-1/3*polylog(2,I*b*exp(I*(d*x^3+c))/(a-(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)
^(1/2)+1/3*polylog(2,I*b*exp(I*(d*x^3+c))/(a+(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)^(1/2)

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Rubi [A]
time = 0.32, 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 = {3460, 3404, 2296, 2221, 2317, 2438} \begin {gather*} -\frac {\text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 d^2 \sqrt {a^2-b^2}}+\frac {\text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^3\right )}}{\sqrt {a^2-b^2}+a}\right )}{3 d^2 \sqrt {a^2-b^2}}-\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 d \sqrt {a^2-b^2}}+\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{\sqrt {a^2-b^2}+a}\right )}{3 d \sqrt {a^2-b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*Sin[c + d*x^3]),x]

[Out]

((-1/3*I)*x^3*Log[1 - (I*b*E^(I*(c + d*x^3)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + ((I/3)*x^3*Log[1 -
(I*b*E^(I*(c + d*x^3)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - PolyLog[2, (I*b*E^(I*(c + d*x^3)))/(a -
Sqrt[a^2 - b^2])]/(3*Sqrt[a^2 - b^2]*d^2) + PolyLog[2, (I*b*E^(I*(c + d*x^3)))/(a + Sqrt[a^2 - b^2])]/(3*Sqrt
[a^2 - b^2]*d^2)

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2317

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 2438

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 3404

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 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^5}{a+b \sin \left (c+d x^3\right )} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x}{a+b \sin (c+d x)} \, dx,x,x^3\right )\\ &=\frac {2}{3} \text {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^3\right )\\ &=-\frac {(2 i b) \text {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^3\right )}{3 \sqrt {a^2-b^2}}+\frac {(2 i b) \text {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^3\right )}{3 \sqrt {a^2-b^2}}\\ &=-\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}+\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}+\frac {i \text {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^3\right )}{3 \sqrt {a^2-b^2} d}-\frac {i \text {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^3\right )}{3 \sqrt {a^2-b^2} d}\\ &=-\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}+\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}+\frac {\text {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^3\right )}\right )}{3 \sqrt {a^2-b^2} d^2}-\frac {\text {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^3\right )}\right )}{3 \sqrt {a^2-b^2} d^2}\\ &=-\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}+\frac {i x^3 \log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d}-\frac {\text {Li}_2\left (\frac {i b e^{i \left (c+d x^3\right )}}{a-\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d^2}+\frac {\text {Li}_2\left (\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d^2}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 188, normalized size = 0.77 \begin {gather*} \frac {-i d x^3 \left (\log \left (1+\frac {i b e^{i \left (c+d x^3\right )}}{-a+\sqrt {a^2-b^2}}\right )-\log \left (1-\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )\right )-\text {Li}_2\left (-\frac {i b e^{i \left (c+d x^3\right )}}{-a+\sqrt {a^2-b^2}}\right )+\text {Li}_2\left (\frac {i b e^{i \left (c+d x^3\right )}}{a+\sqrt {a^2-b^2}}\right )}{3 \sqrt {a^2-b^2} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*Sin[c + d*x^3]),x]

[Out]

((-I)*d*x^3*(Log[1 + (I*b*E^(I*(c + d*x^3)))/(-a + Sqrt[a^2 - b^2])] - Log[1 - (I*b*E^(I*(c + d*x^3)))/(a + Sq
rt[a^2 - b^2])]) - PolyLog[2, ((-I)*b*E^(I*(c + d*x^3)))/(-a + Sqrt[a^2 - b^2])] + PolyLog[2, (I*b*E^(I*(c + d
*x^3)))/(a + Sqrt[a^2 - b^2])])/(3*Sqrt[a^2 - b^2]*d^2)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{5}}{a +b \sin \left (d \,x^{3}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a+b*sin(d*x^3+c)),x)

[Out]

int(x^5/(a+b*sin(d*x^3+c)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b*sin(d*x^3+c)),x, algorithm="maxima")

[Out]

integrate(x^5/(b*sin(d*x^3 + c) + a), x)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1041 vs. \(2 (199) = 398\).
time = 0.56, size = 1041, normalized size = 4.25 \begin {gather*} -\frac {b c \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x^{3} + c\right ) + 2 i \, b \sin \left (d x^{3} + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) + b c \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x^{3} + c\right ) - 2 i \, b \sin \left (d x^{3} + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - b c \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x^{3} + c\right ) + 2 i \, b \sin \left (d x^{3} + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - b c \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x^{3} + c\right ) - 2 i \, b \sin \left (d x^{3} + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - i \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) + {\left (b \cos \left (d x^{3} + c\right ) + i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) - {\left (b \cos \left (d x^{3} + c\right ) + i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) + {\left (b \cos \left (d x^{3} + c\right ) - i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - i \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) - {\left (b \cos \left (d x^{3} + c\right ) - i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (b d x^{3} + b c\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) + {\left (b \cos \left (d x^{3} + c\right ) + i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - {\left (b d x^{3} + b c\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) - {\left (b \cos \left (d x^{3} + c\right ) + i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d x^{3} + b c\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) + {\left (b \cos \left (d x^{3} + c\right ) - i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - {\left (b d x^{3} + b c\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (d x^{3} + c\right ) - a \sin \left (d x^{3} + c\right ) - {\left (b \cos \left (d x^{3} + c\right ) - i \, b \sin \left (d x^{3} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right )}{6 \, {\left (a^{2} - b^{2}\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b*sin(d*x^3+c)),x, algorithm="fricas")

[Out]

-1/6*(b*c*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x^3 + c) + 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) +
2*I*a) + b*c*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x^3 + c) - 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2)
- 2*I*a) - b*c*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x^3 + c) + 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2 - b^2)/
b^2) + 2*I*a) - b*c*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x^3 + c) - 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2 - b
^2)/b^2) - 2*I*a) - I*b*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x^3 + c) - a*sin(d*x^3 + c) + (b*cos(d*x^3 + c
) + I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + I*b*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x^3 +
c) - a*sin(d*x^3 + c) - (b*cos(d*x^3 + c) + I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + I*b*sqrt
(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x^3 + c) - a*sin(d*x^3 + c) + (b*cos(d*x^3 + c) - I*b*sin(d*x^3 + c))*sqr
t(-(a^2 - b^2)/b^2) - b)/b + 1) - I*b*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x^3 + c) - a*sin(d*x^3 + c) - (
b*cos(d*x^3 + c) - I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + (b*d*x^3 + b*c)*sqrt(-(a^2 - b^2)/
b^2)*log(-(I*a*cos(d*x^3 + c) - a*sin(d*x^3 + c) + (b*cos(d*x^3 + c) + I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b
^2) - b)/b) - (b*d*x^3 + b*c)*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x^3 + c) - a*sin(d*x^3 + c) - (b*cos(d*x^
3 + c) + I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (b*d*x^3 + b*c)*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*
a*cos(d*x^3 + c) - a*sin(d*x^3 + c) + (b*cos(d*x^3 + c) - I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) -
(b*d*x^3 + b*c)*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x^3 + c) - a*sin(d*x^3 + c) - (b*cos(d*x^3 + c) - I*b
*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b))/((a^2 - b^2)*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{a + b \sin {\left (c + d x^{3} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(a+b*sin(d*x**3+c)),x)

[Out]

Integral(x**5/(a + b*sin(c + d*x**3)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b*sin(d*x^3+c)),x, algorithm="giac")

[Out]

integrate(x^5/(b*sin(d*x^3 + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^5}{a+b\,\sin \left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*sin(c + d*x^3)),x)

[Out]

int(x^5/(a + b*sin(c + d*x^3)), x)

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