\(\int \frac {(b+a x^3) \sqrt {x+x^4}}{-d+c x^3} \, dx\) [1839]

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {\left (b+a x^3\right ) \sqrt {x+x^4}}{-d+c x^3} \, dx=\frac {a x \sqrt {x+x^4}}{3 c}+\frac {2 \sqrt {-c-d} (b c+a d) \arctan \left (\frac {\sqrt {-c-d} x \sqrt {x+x^4}}{\sqrt {d} (1+x) \left (1-x+x^2\right )}\right )}{3 c^2 \sqrt {d}}+\frac {(a c+2 b c+2 a d) \text {arctanh}\left (\frac {x^2}{\sqrt {x+x^4}}\right )}{3 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.15, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2081, 595, 598, 335, 281, 221, 477, 476, 385, 214} \[ \int \frac {\left (b+a x^3\right ) \sqrt {x+x^4}}{-d+c x^3} \, dx=\frac {\sqrt {x^4+x} \text {arcsinh}\left (x^{3/2}\right ) (a (c+2 d)+2 b c)}{3 c^2 \sqrt {x^3+1} \sqrt {x}}-\frac {2 \sqrt {x^4+x} \sqrt {c+d} (a d+b c) \text {arctanh}\left (\frac {x^{3/2} \sqrt {c+d}}{\sqrt {d} \sqrt {x^3+1}}\right )}{3 c^2 \sqrt {d} \sqrt {x^3+1} \sqrt {x}}+\frac {a \sqrt {x^4+x} x}{3 c} \]

[In]

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

[Out]

(a*x*Sqrt[x + x^4])/(3*c) + ((2*b*c + a*(c + 2*d))*Sqrt[x + x^4]*ArcSinh[x^(3/2)])/(3*c^2*Sqrt[x]*Sqrt[1 + x^3
]) - (2*Sqrt[c + d]*(b*c + a*d)*Sqrt[x + x^4]*ArcTanh[(Sqrt[c + d]*x^(3/2))/(Sqrt[d]*Sqrt[1 + x^3])])/(3*c^2*S
qrt[d]*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 281

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 595

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*g*(m + n*(p + q + 1) + 1))), x] + Dis
t[1/(b*(m + n*(p + q + 1) + 1)), Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*((b*e - a*f)*(m + 1) + b
*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {x+x^4} \int \frac {\sqrt {x} \sqrt {1+x^3} \left (b+a x^3\right )}{-d+c x^3} \, dx}{\sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {\sqrt {x+x^4} \int \frac {\sqrt {x} \left (\frac {3}{2} (2 b c+a d)+\frac {3}{2} (2 b c+a (c+2 d)) x^3\right )}{\sqrt {1+x^3} \left (-d+c x^3\right )} \, dx}{3 c \sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {\sqrt {x+x^4} \int \left (\frac {3 (2 b c+a (c+2 d)) \sqrt {x}}{2 c \sqrt {1+x^3}}+\frac {\left (\frac {3}{2} c (2 b c+a d)+\frac {3}{2} d (2 b c+a (c+2 d))\right ) \sqrt {x}}{c \sqrt {1+x^3} \left (-d+c x^3\right )}\right ) \, dx}{3 c \sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {\left ((c+d) (b c+a d) \sqrt {x+x^4}\right ) \int \frac {\sqrt {x}}{\sqrt {1+x^3} \left (-d+c x^3\right )} \, dx}{c^2 \sqrt {x} \sqrt {1+x^3}}+\frac {\left ((2 b c+a (c+2 d)) \sqrt {x+x^4}\right ) \int \frac {\sqrt {x}}{\sqrt {1+x^3}} \, dx}{2 c^2 \sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {\left (2 (c+d) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^6} \left (-d+c x^6\right )} \, dx,x,\sqrt {x}\right )}{c^2 \sqrt {x} \sqrt {1+x^3}}+\frac {\left ((2 b c+a (c+2 d)) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{c^2 \sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {\left (2 (c+d) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (-d+c x^2\right )} \, dx,x,x^{3/2}\right )}{3 c^2 \sqrt {x} \sqrt {1+x^3}}+\frac {\left ((2 b c+a (c+2 d)) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^{3/2}\right )}{3 c^2 \sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {(2 b c+a (c+2 d)) \sqrt {x+x^4} \text {arcsinh}\left (x^{3/2}\right )}{3 c^2 \sqrt {x} \sqrt {1+x^3}}+\frac {\left (2 (c+d) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{-d-(-c-d) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {1+x^3}}\right )}{3 c^2 \sqrt {x} \sqrt {1+x^3}} \\ & = \frac {a x \sqrt {x+x^4}}{3 c}+\frac {(2 b c+a (c+2 d)) \sqrt {x+x^4} \text {arcsinh}\left (x^{3/2}\right )}{3 c^2 \sqrt {x} \sqrt {1+x^3}}-\frac {2 \sqrt {c+d} (b c+a d) \sqrt {x+x^4} \text {arctanh}\left (\frac {\sqrt {c+d} x^{3/2}}{\sqrt {d} \sqrt {1+x^3}}\right )}{3 c^2 \sqrt {d} \sqrt {x} \sqrt {1+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.15 \[ \int \frac {\left (b+a x^3\right ) \sqrt {x+x^4}}{-d+c x^3} \, dx=\frac {\sqrt {x+x^4} \left (2 \sqrt {c+d} (b c+a d) \text {arctanh}\left (\frac {d-c x^{3/2} \left (x^{3/2}+\sqrt {1+x^3}\right )}{\sqrt {d} \sqrt {c+d}}\right )+\sqrt {d} \left (a c x^{3/2} \sqrt {1+x^3}+(a c+2 b c+2 a d) \log \left (x^{3/2}+\sqrt {1+x^3}\right )\right )\right )}{3 c^2 \sqrt {d} \sqrt {x} \sqrt {1+x^3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {a \,x^{2} \left (x^{3}+1\right )}{3 c \sqrt {x \left (x^{3}+1\right )}}+\frac {-\frac {\left (a c +2 a d +2 b c \right ) \ln \left (2 x^{3}-2 x \sqrt {x^{4}+x}+1\right )}{3 c}-\frac {4 \left (a c d +a \,d^{2}+b \,c^{2}+b c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{4}+x}\, d}{x^{2} \sqrt {\left (c +d \right ) d}}\right )}{3 c \sqrt {\left (c +d \right ) d}}}{2 c}\)	\(117\)
pseudoelliptic	\(-\frac {2 \left (-\frac {\sqrt {x^{4}+x}\, \sqrt {\left (c +d \right ) d}\, a c x}{2}+\frac {\left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )\right ) \left (\left (c +2 d \right ) a +2 b c \right ) \sqrt {\left (c +d \right ) d}}{4}+\left (c +d \right ) \left (a d +b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{4}+x}\, d}{x^{2} \sqrt {\left (c +d \right ) d}}\right )\right )}{3 \sqrt {\left (c +d \right ) d}\, c^{2}}\)	\(122\)
default	\(\frac {a \left (\frac {x \sqrt {x^{4}+x}}{3}+\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}-\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}\right )}{c}+\frac {\left (a d +b c \right ) \left (\ln \left (\frac {x^{2}+\sqrt {x \left (x^{3}+1\right )}}{x^{2}}\right )-\frac {2 \left (c +d \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{3}+1\right )}\, d}{x^{2} \sqrt {\left (c +d \right ) d}}\right )}{\sqrt {\left (c +d \right ) d}}-\ln \left (\frac {-x^{2}+\sqrt {x \left (x^{3}+1\right )}}{x^{2}}\right )\right )}{3 c^{2}}\)	\(143\)
elliptic	\(\text {Expression too large to display}\)	\(701\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 2.48 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.13 \[ \int \frac {\left (b+a x^3\right ) \sqrt {x+x^4}}{-d+c x^3} \, dx=\left [\frac {2 \, \sqrt {x^{4} + x} a c x + {\left (b c + a d\right )} \sqrt {\frac {c + d}{d}} \log \left (-\frac {{\left (c^{2} + 8 \, c d + 8 \, d^{2}\right )} x^{6} + 2 \, {\left (3 \, c d + 4 \, d^{2}\right )} x^{3} + d^{2} - 4 \, {\left ({\left (c d + 2 \, d^{2}\right )} x^{4} + d^{2} x\right )} \sqrt {x^{4} + x} \sqrt {\frac {c + d}{d}}}{c^{2} x^{6} - 2 \, c d x^{3} + d^{2}}\right ) + {\left ({\left (a + 2 \, b\right )} c + 2 \, a d\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right )}{6 \, c^{2}}, \frac {2 \, \sqrt {x^{4} + x} a c x + 2 \, {\left (b c + a d\right )} \sqrt {-\frac {c + d}{d}} \arctan \left (\frac {2 \, \sqrt {x^{4} + x} d x \sqrt {-\frac {c + d}{d}}}{{\left (c + 2 \, d\right )} x^{3} + d}\right ) + {\left ({\left (a + 2 \, b\right )} c + 2 \, a d\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right )}{6 \, c^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02 \[ \int \frac {\left (b+a x^3\right ) \sqrt {x+x^4}}{-d+c x^3} \, dx=\frac {\sqrt {x^{4} + x} a x}{3 \, c} + \frac {{\left (a c + 2 \, b c + 2 \, a d\right )} \log \left (\sqrt {\frac {1}{x^{3}} + 1} + 1\right )}{6 \, c^{2}} - \frac {{\left (a c + 2 \, b c + 2 \, a d\right )} \log \left ({\left | \sqrt {\frac {1}{x^{3}} + 1} - 1 \right |}\right )}{6 \, c^{2}} + \frac {2 \, {\left (b c^{2} + a c d + b c d + a d^{2}\right )} \arctan \left (\frac {d \sqrt {\frac {1}{x^{3}} + 1}}{\sqrt {-c d - d^{2}}}\right )}{3 \, \sqrt {-c d - d^{2}} c^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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