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

3.17.11.1 Optimal result

Integrand size = 24, antiderivative size = 110 \[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=\frac {x \sqrt {x+x^4}}{3 a}+\frac {2 \sqrt {-a-b} \sqrt {b} \arctan \left (\frac {\sqrt {-a-b} x \sqrt {x+x^4}}{\sqrt {b} (1+x) \left (1-x+x^2\right )}\right )}{3 a^2}+\frac {(a+2 b) \text {arctanh}\left (\frac {x^2}{\sqrt {x+x^4}}\right )}{3 a^2} \]

output

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
 

3.17.11.2 Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.11 \[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=\frac {\sqrt {x+x^4} \left (a x^{3/2} \sqrt {1+x^3}+2 \sqrt {b} \sqrt {a+b} \text {arctanh}\left (\frac {b-a x^{3/2} \left (x^{3/2}+\sqrt {1+x^3}\right )}{\sqrt {b} \sqrt {a+b}}\right )+(a+2 b) \log \left (x^{3/2}+\sqrt {1+x^3}\right )\right )}{3 a^2 \sqrt {x} \sqrt {1+x^3}} \]

input

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

output

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

3.17.11.3 Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1948, 25, 966, 965, 380, 398, 222, 291, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^3 \sqrt {x^4+x}}{a x^3-b} \, dx\)

\(\Big \downarrow \) 1948

\(\displaystyle \frac {\sqrt {x^4+x} \int -\frac {x^{7/2} \sqrt {x^3+1}}{b-a x^3}dx}{\sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x^4+x} \int \frac {x^{7/2} \sqrt {x^3+1}}{b-a x^3}dx}{\sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 966

\(\displaystyle -\frac {2 \sqrt {x^4+x} \int \frac {x^4 \sqrt {x^3+1}}{b-a x^3}d\sqrt {x}}{\sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 965

\(\displaystyle -\frac {2 \sqrt {x^4+x} \int \frac {x \sqrt {x+1}}{b-a x}dx^{3/2}}{3 \sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 380

\(\displaystyle -\frac {2 \sqrt {x^4+x} \left (\frac {\int \frac {b+(a+2 b) x}{\sqrt {x+1} (b-a x)}dx^{3/2}}{2 a}-\frac {x^{3/2} \sqrt {x+1}}{2 a}\right )}{3 \sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 398

\(\displaystyle -\frac {2 \sqrt {x^4+x} \left (\frac {\frac {2 b (a+b) \int \frac {1}{\sqrt {x+1} (b-a x)}dx^{3/2}}{a}-\frac {(a+2 b) \int \frac {1}{\sqrt {x+1}}dx^{3/2}}{a}}{2 a}-\frac {x^{3/2} \sqrt {x+1}}{2 a}\right )}{3 \sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 222

\(\displaystyle -\frac {2 \sqrt {x^4+x} \left (\frac {\frac {2 b (a+b) \int \frac {1}{\sqrt {x+1} (b-a x)}dx^{3/2}}{a}-\frac {(a+2 b) \text {arcsinh}\left (x^{3/2}\right )}{a}}{2 a}-\frac {x^{3/2} \sqrt {x+1}}{2 a}\right )}{3 \sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {2 \sqrt {x^4+x} \left (\frac {\frac {2 b (a+b) \int \frac {1}{b-(a+b) x}d\frac {x^{3/2}}{\sqrt {x+1}}}{a}-\frac {(a+2 b) \text {arcsinh}\left (x^{3/2}\right )}{a}}{2 a}-\frac {x^{3/2} \sqrt {x+1}}{2 a}\right )}{3 \sqrt {x} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \sqrt {x^4+x} \left (\frac {\frac {2 \sqrt {b} \sqrt {a+b} \text {arctanh}\left (\frac {x^{3/2} \sqrt {a+b}}{\sqrt {b} \sqrt {x+1}}\right )}{a}-\frac {(a+2 b) \text {arcsinh}\left (x^{3/2}\right )}{a}}{2 a}-\frac {x^{3/2} \sqrt {x+1}}{2 a}\right )}{3 \sqrt {x} \sqrt {x^3+1}}\)

input

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

output

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

3.17.11.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 221

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 222

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 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 380

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b* 
(m + 2*(p + q) + 1))), x] - Simp[e^2/(b*(m + 2*(p + q) + 1))   Int[(e*x)^(m 
 - 2)*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[a*c*(m - 1) + (a*d*(m - 1) - 2 
*q*(b*c - a*d))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c 
- a*d, 0] && GtQ[q, 0] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, 
 q, x]
 

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 965

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

rule 966

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> With[{k = Denominator[m]}, Simp[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] && IntegerQ[p]
 

rule 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[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])
 

3.17.11.4 Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {x^{2} \left (x^{3}+1\right )}{3 a \sqrt {x \left (x^{3}+1\right )}}+\frac {-\frac {\left (a +2 b \right ) \ln \left (2 x^{3}-2 x \sqrt {x^{4}+x}+1\right )}{3 a}-\frac {4 \left (a +b \right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {x^{4}+x}\, b}{x^{2} \sqrt {\left (a +b \right ) b}}\right )}{3 a \sqrt {\left (a +b \right ) b}}}{2 a}\)	\(94\)
pseudoelliptic	\(-\frac {-2 \sqrt {\left (a +b \right ) b}\, \sqrt {x^{4}+x}\, a x +\left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )\right ) \left (a +2 b \right ) \sqrt {\left (a +b \right ) b}+4 \left (a +b \right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {x^{4}+x}\, b}{x^{2} \sqrt {\left (a +b \right ) b}}\right )}{6 \sqrt {\left (a +b \right ) b}\, a^{2}}\)	\(108\)
default	\(\frac {\frac {x \sqrt {x^{4}+x}}{3}+\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}-\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}}{a}+\frac {b \left (\ln \left (\frac {x^{2}+\sqrt {x \left (x^{3}+1\right )}}{x^{2}}\right )-\frac {2 \left (a +b \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{3}+1\right )}\, b}{x^{2} \sqrt {\left (a +b \right ) b}}\right )}{\sqrt {\left (a +b \right ) b}}-\ln \left (\frac {-x^{2}+\sqrt {x \left (x^{3}+1\right )}}{x^{2}}\right )\right )}{3 a^{2}}\)	\(136\)
elliptic	\(\text {Expression too large to display}\)	\(665\)

input

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

output

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

3.17.11.5 Fricas [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.12 \[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=\left [\frac {2 \, \sqrt {x^{4} + x} a x + {\left (a + 2 \, b\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right ) + \sqrt {a b + b^{2}} \log \left (-\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} x^{6} + 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} x^{3} - 4 \, {\left ({\left (a + 2 \, b\right )} x^{4} + b x\right )} \sqrt {x^{4} + x} \sqrt {a b + b^{2}} + b^{2}}{a^{2} x^{6} - 2 \, a b x^{3} + b^{2}}\right )}{6 \, a^{2}}, \frac {2 \, \sqrt {x^{4} + x} a x + {\left (a + 2 \, b\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right ) + 2 \, \sqrt {-a b - b^{2}} \arctan \left (\frac {2 \, \sqrt {x^{4} + x} \sqrt {-a b - b^{2}} x}{{\left (a + 2 \, b\right )} x^{3} + b}\right )}{6 \, a^{2}}\right ] \]

input

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

output

[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]
 

3.17.11.6 Sympy [F]

\[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=\int \frac {x^{3} \sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )}}{a x^{3} - b}\, dx \]

input

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

output

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

3.17.11.7 Maxima [F]

\[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=\int { \frac {\sqrt {x^{4} + x} x^{3}}{a x^{3} - b} \,d x } \]

input

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

output

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

3.17.11.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=\frac {\sqrt {x^{4} + x} x}{3 \, a} + \frac {{\left (a + 2 \, b\right )} \log \left (\sqrt {\frac {1}{x^{3}} + 1} + 1\right )}{6 \, a^{2}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | \sqrt {\frac {1}{x^{3}} + 1} - 1 \right |}\right )}{6 \, a^{2}} + \frac {2 \, {\left (a b + b^{2}\right )} \arctan \left (\frac {b \sqrt {\frac {1}{x^{3}} + 1}}{\sqrt {-a b - b^{2}}}\right )}{3 \, \sqrt {-a b - b^{2}} a^{2}} \]

input

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

output

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)
 

3.17.11.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \sqrt {x+x^4}}{-b+a x^3} \, dx=-\int \frac {x^3\,\sqrt {x^4+x}}{b-a\,x^3} \,d x \]

input

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

output

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