3.17.11 \(\int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx\) [1611]

Optimal. Leaf size=110 \[ \frac {x \sqrt {x+x^4}}{3 a}+\frac {2 \sqrt {-a-b} \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {-a-b} x \sqrt {x+x^4}}{\sqrt {b} (1+x) \left (1-x+x^2\right )}\right )}{3 a^2}+\frac {(a+2 b) \tanh ^{-1}\left (\frac {x^2}{\sqrt {x+x^4}}\right )}{3 a^2} \]

[Out]

1/3*x*(x^4+x)^(1/2)/a+2/3*(-a-b)^(1/2)*b^(1/2)*arctan((-a-b)^(1/2)*x*(x^4+x)^(1/2)/b^(1/2)/(1+x)/(x^2-x+1))/a^
2+1/3*(a+2*b)*arctanh(x^2/(x^4+x)^(1/2))/a^2

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Rubi [A]
time = 0.16, antiderivative size = 129, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2067, 477, 476, 489, 537, 221, 385, 214} \begin {gather*} \frac {\sqrt {x^4+x} (a+2 b) \sinh ^{-1}\left (x^{3/2}\right )}{3 a^2 \sqrt {x^3+1} \sqrt {x}}-\frac {2 \sqrt {b} \sqrt {x^4+x} \sqrt {a+b} \tanh ^{-1}\left (\frac {x^{3/2} \sqrt {a+b}}{\sqrt {b} \sqrt {x^3+1}}\right )}{3 a^2 \sqrt {x^3+1} \sqrt {x}}+\frac {\sqrt {x^4+x} x}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*Sqrt[x + x^4])/(-b + a*x^3),x]

[Out]

(x*Sqrt[x + x^4])/(3*a) + ((a + 2*b)*Sqrt[x + x^4]*ArcSinh[x^(3/2)])/(3*a^2*Sqrt[x]*Sqrt[1 + x^3]) - (2*Sqrt[b
]*Sqrt[a + b]*Sqrt[x + x^4]*ArcTanh[(Sqrt[a + b]*x^(3/2))/(Sqrt[b]*Sqrt[1 + x^3])])/(3*a^2*Sqrt[x]*Sqrt[1 + x^
3])

Rule 214

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

Rule 221

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 476

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 489

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(m + n*(p + q) + 1))), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 2067

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[e^IntPart[m]*(e*x)^FracPart[m]*((a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p])), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx &=\frac {\sqrt {x+x^4} \int \frac {x^{7/2} \sqrt {1+x^3}}{-b+a x^3} \, dx}{\sqrt {x} \sqrt {1+x^3}}\\ &=\frac {\left (2 \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {x^8 \sqrt {1+x^6}}{-b+a x^6} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+x^3}}\\ &=\frac {\left (2 \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {1+x^2}}{-b+a x^2} \, dx,x,x^{3/2}\right )}{3 \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {x \sqrt {x+x^4}}{3 a}-\frac {\sqrt {x+x^4} \text {Subst}\left (\int \frac {-b+(-a-2 b) x^2}{\sqrt {1+x^2} \left (-b+a x^2\right )} \, dx,x,x^{3/2}\right )}{3 a \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {x \sqrt {x+x^4}}{3 a}-\frac {\left ((-a-2 b) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^{3/2}\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}}+\frac {\left (2 b (a+b) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (-b+a x^2\right )} \, dx,x,x^{3/2}\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {x \sqrt {x+x^4}}{3 a}+\frac {(a+2 b) \sqrt {x+x^4} \sinh ^{-1}\left (x^{3/2}\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}}+\frac {\left (2 b (a+b) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{-b-(-a-b) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {1+x^3}}\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {x \sqrt {x+x^4}}{3 a}+\frac {(a+2 b) \sqrt {x+x^4} \sinh ^{-1}\left (x^{3/2}\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}}-\frac {2 \sqrt {b} \sqrt {a+b} \sqrt {x+x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b} x^{3/2}}{\sqrt {b} \sqrt {1+x^3}}\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 122, normalized size = 1.11 \begin {gather*} \frac {\sqrt {x+x^4} \left (a x^{3/2} \sqrt {1+x^3}+(a+2 b) \tanh ^{-1}\left (\frac {x^{3/2}}{\sqrt {1+x^3}}\right )+2 \sqrt {b} \sqrt {a+b} \tanh ^{-1}\left (\frac {b-a x^{3/2} \left (x^{3/2}+\sqrt {1+x^3}\right )}{\sqrt {b} \sqrt {a+b}}\right )\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sqrt[x + x^4])/(-b + a*x^3),x]

[Out]

(Sqrt[x + x^4]*(a*x^(3/2)*Sqrt[1 + x^3] + (a + 2*b)*ArcTanh[x^(3/2)/Sqrt[1 + x^3]] + 2*Sqrt[b]*Sqrt[a + b]*Arc
Tanh[(b - a*x^(3/2)*(x^(3/2) + Sqrt[1 + x^3]))/(Sqrt[b]*Sqrt[a + b])]))/(3*a^2*Sqrt[x]*Sqrt[1 + x^3])

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.46, size = 963, normalized size = 8.75 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(x^4+x)^(1/2)/(a*x^3-b),x,method=_RETURNVERBOSE)

[Out]

1/a*(1/3*x*(x^4+x)^(1/2)-(-1/2-1/2*I*3^(1/2))*((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*
(-(x-1/2+1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2))/(1+x))^(1/2)*(-(x-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(1+x))^(1
/2)/(3/2+1/2*I*3^(1/2))/(x*(1+x)*(x-1/2+1/2*I*3^(1/2))*(x-1/2-1/2*I*3^(1/2)))^(1/2)*(-EllipticF(((3/2+1/2*I*3^
(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/
2-1/2*I*3^(1/2)))^(1/2))+EllipticPi(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),(1/2+1/2*I*3^(1/2)
)/(3/2+1/2*I*3^(1/2)),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1
/2))))+b/a*(-2/a*(-1/2-1/2*I*3^(1/2))*((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*(-(x-1/2
+1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2))/(1+x))^(1/2)*(-(x-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)/(3/2
+1/2*I*3^(1/2))/(x*(1+x)*(x-1/2+1/2*I*3^(1/2))*(x-1/2-1/2*I*3^(1/2)))^(1/2)*(-EllipticF(((3/2+1/2*I*3^(1/2))*x
/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*
3^(1/2)))^(1/2))+EllipticPi(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),(1/2+1/2*I*3^(1/2))/(3/2+1
/2*I*3^(1/2)),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)))-2/
3/a*4^(1/2)*sum((-a-b)/_alpha*(1+x)^2*(_alpha^2-_alpha+1)/(a+b)*(-1-I*3^(1/2))*(x/(1+x)*(3+I*3^(1/2))/(1+I*3^(
1/2)))^(1/2)*(-1/(1+x)*(I*3^(1/2)+2*x-1)/(1-I*3^(1/2)))^(1/2)*(-1/(1+x)*(-I*3^(1/2)+2*x-1)/(1+I*3^(1/2)))^(1/2
)/(3+I*3^(1/2))/(x*(1+x)*(I*3^(1/2)+2*x-1)*(-I*3^(1/2)+2*x-1))^(1/2)*(EllipticF(((3/2+1/2*I*3^(1/2))*x/(1/2+1/
2*I*3^(1/2))/(1+x))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2))
)^(1/2))+_alpha^2*a/b*EllipticPi(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),1/6*(I*_alpha^2*3^(1/
2)*a+3*_alpha^2*a+I*3^(1/2)*b+3*b)/b,((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2
*I*3^(1/2)))^(1/2))),_alpha=RootOf(_Z^3*a-b)))

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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^3*(x^4+x)^(1/2)/(a*x^3-b),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 + x)*x^3/(a*x^3 - b), x)

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Fricas [A]
time = 1.27, size = 233, normalized size = 2.12 \begin {gather*} \left [\frac {2 \, \sqrt {x^{4} + x} a x + {\left (a + 2 \, b\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right ) + \sqrt {a b + b^{2}} \log \left (-\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} x^{6} + 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} x^{3} - 4 \, {\left ({\left (a + 2 \, b\right )} x^{4} + b x\right )} \sqrt {x^{4} + x} \sqrt {a b + b^{2}} + b^{2}}{a^{2} x^{6} - 2 \, a b x^{3} + b^{2}}\right )}{6 \, a^{2}}, \frac {2 \, \sqrt {x^{4} + x} a x + {\left (a + 2 \, b\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right ) + 2 \, \sqrt {-a b - b^{2}} \arctan \left (\frac {2 \, \sqrt {x^{4} + x} \sqrt {-a b - b^{2}} x}{{\left (a + 2 \, b\right )} x^{3} + b}\right )}{6 \, a^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(x^4+x)^(1/2)/(a*x^3-b),x, algorithm="fricas")

[Out]

[1/6*(2*sqrt(x^4 + x)*a*x + (a + 2*b)*log(-2*x^3 - 2*sqrt(x^4 + x)*x - 1) + sqrt(a*b + b^2)*log(-((a^2 + 8*a*b
+ 8*b^2)*x^6 + 2*(3*a*b + 4*b^2)*x^3 - 4*((a + 2*b)*x^4 + b*x)*sqrt(x^4 + x)*sqrt(a*b + b^2) + b^2)/(a^2*x^6
- 2*a*b*x^3 + b^2)))/a^2, 1/6*(2*sqrt(x^4 + x)*a*x + (a + 2*b)*log(-2*x^3 - 2*sqrt(x^4 + x)*x - 1) + 2*sqrt(-a
*b - b^2)*arctan(2*sqrt(x^4 + x)*sqrt(-a*b - b^2)*x/((a + 2*b)*x^3 + b)))/a^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )}}{a x^{3} - b}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(x**4+x)**(1/2)/(a*x**3-b),x)

[Out]

Integral(x**3*sqrt(x*(x + 1)*(x**2 - x + 1))/(a*x**3 - b), x)

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Giac [A]
time = 0.48, size = 101, normalized size = 0.92 \begin {gather*} \frac {\sqrt {x^{4} + x} x}{3 \, a} + \frac {{\left (a + 2 \, b\right )} \log \left (\sqrt {\frac {1}{x^{3}} + 1} + 1\right )}{6 \, a^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | \sqrt {\frac {1}{x^{3}} + 1} - 1 \right |}\right )}{6 \, a^{2}} + \frac {2 \, {\left (a b + b^{2}\right )} \arctan \left (\frac {b \sqrt {\frac {1}{x^{3}} + 1}}{\sqrt {-a b - b^{2}}}\right )}{3 \, \sqrt {-a b - b^{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(x^4+x)^(1/2)/(a*x^3-b),x, algorithm="giac")

[Out]

1/3*sqrt(x^4 + x)*x/a + 1/6*(a + 2*b)*log(sqrt(1/x^3 + 1) + 1)/a^2 - 1/6*(a + 2*b)*log(abs(sqrt(1/x^3 + 1) - 1
))/a^2 + 2/3*(a*b + b^2)*arctan(b*sqrt(1/x^3 + 1)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x^3\,\sqrt {x^4+x}}{b-a\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*(x + x^4)^(1/2))/(b - a*x^3),x)

[Out]

-int((x^3*(x + x^4)^(1/2))/(b - a*x^3), x)

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