\(\int \frac {(b+a x^3) \sqrt {-x+x^4}}{-d+c x^3} \, dx\) [1847]

Optimal result

Integrand size = 30, antiderivative size = 126 \[ \int \frac {\left (b+a x^3\right ) \sqrt {-x+x^4}}{-d+c x^3} \, dx=\frac {a x \sqrt {-x+x^4}}{3 c}+\frac {2 \sqrt {c-d} (b c+a d) \arctan \left (\frac {\sqrt {c-d} x \sqrt {-x+x^4}}{\sqrt {d} (-1+x) \left (1+x+x^2\right )}\right )}{3 c^2 \sqrt {d}}+\frac {(-a c+2 b c+2 a d) \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right )}{3 c^2} \]

[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 = 0.24 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.30, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {2081, 595, 598, 335, 281, 223, 212, 477, 476, 385, 211} \[ \int \frac {\left (b+a x^3\right ) \sqrt {-x+x^4}}{-d+c x^3} \, dx=\frac {2 \sqrt {x^4-x} \sqrt {c-d} (a d+b c) \arctan \left (\frac {x^{3/2} \sqrt {c-d}}{\sqrt {d} \sqrt {x^3-1}}\right )}{3 c^2 \sqrt {d} \sqrt {x^3-1} \sqrt {x}}-\frac {\sqrt {x^4-x} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x^3-1}}\right ) (a c-2 a d-2 b c)}{3 c^2 \sqrt {x^3-1} \sqrt {x}}+\frac {a \sqrt {x^4-x} x}{3 c} \]

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

Rule 211

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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

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} (a c-2 b c-2 a 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 (a c-2 b c-2 a d) \sqrt {x}}{2 c \sqrt {-1+x^3}}+\frac {\left (-\frac {3}{2} d (a c-2 b c-2 a d)-\frac {3}{2} c (2 b c+a 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 ((a c-2 b c-2 a 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 {\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 {a x \sqrt {-x+x^4}}{3 c}-\frac {\left ((a c-2 b c-2 a 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 {\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 {a x \sqrt {-x+x^4}}{3 c}-\frac {\left ((a c-2 b c-2 a 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 {\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 {a x \sqrt {-x+x^4}}{3 c}-\frac {\left ((a c-2 b c-2 a d) \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {-1+x^3}}\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 \sqrt {c-d} (b c+a d) \sqrt {-x+x^4} \arctan \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}}+\frac {(2 b c-a (c-2 d)) \sqrt {-x+x^4} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {-1+x^3}}\right )}{3 c^2 \sqrt {x} \sqrt {-1+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.20 \[ \int \frac {\left (b+a x^3\right ) \sqrt {-x+x^4}}{-d+c x^3} \, dx=\frac {\sqrt {x} \sqrt {-1+x^3} \left (-2 \sqrt {c-d} (b c+a d) \arctan \left (\frac {-d+c x^3+c x^{3/2} \sqrt {-1+x^3}}{\sqrt {c-d} \sqrt {d}}\right )+\sqrt {d} \left (a c x^{3/2} \sqrt {-1+x^3}+(-a c+2 b c+2 a d) \log \left (x^{3/2}+\sqrt {-1+x^3}\right )\right )\right )}{3 c^2 \sqrt {d} \sqrt {x \left (-1+x^3\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.01

method	result	size
risch	\(\frac {a \,x^{2} \left (x^{3}-1\right )}{3 c \sqrt {x \left (x^{3}-1\right )}}-\frac {-\frac {\left (a c -2 a d -2 b c \right ) \ln \left (2 x^{3}-2 x \sqrt {x^{4}-x}-1\right )}{3 c}+\frac {4 \left (a c d -a \,d^{2}+b \,c^{2}-b c d \right ) \arctan \left (\frac {\sqrt {x^{4}-x}\, d}{x^{2} \sqrt {\left (c -d \right ) d}}\right )}{3 c \sqrt {\left (c -d \right ) d}}}{2 c}\)	\(127\)
pseudoelliptic	\(-\frac {2 \left (-\frac {\sqrt {\left (c -d \right ) d}\, \sqrt {x^{4}-x}\, a c x}{2}-\frac {\left (\left (-2 d +c \right ) a -2 b c \right ) \left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right ) \sqrt {\left (c -d \right ) d}}{4}+\left (c -d \right ) \left (a d +b c \right ) \arctan \left (\frac {\sqrt {x^{4}-x}\, d}{x^{2} \sqrt {\left (c -d \right ) d}}\right )\right )}{3 \sqrt {\left (c -d \right ) d}\, c^{2}}\)	\(140\)
default	\(\frac {a \left (\frac {x \sqrt {x^{4}-x}}{3}+\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}-\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}\right )}{c}-\frac {\left (a d +b c \right ) \left (\left (2 c -2 d \right ) \arctan \left (\frac {\sqrt {x^{4}-x}\, d}{x^{2} \sqrt {\left (c -d \right ) d}}\right )+\left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right ) \sqrt {\left (c -d \right ) d}\right )}{3 c^{2} \sqrt {\left (c -d \right ) d}}\)	\(167\)
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)*arctan((x^4-x)^(1/2)/x^2*d/((c-d)*d)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 2.41 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.29 \[ \int \frac {\left (b+a x^3\right ) \sqrt {-x+x^4}}{-d+c x^3} \, dx=\left [\frac {2 \, \sqrt {x^{4} - x} a c x + {\left (b c + a d\right )} \sqrt {-\frac {c - d}{d}} \log \left (-\frac {{\left (c^{2} - 8 \, c d + 8 \, d^{2}\right )} x^{6} + 2 \, {\left (3 \, c d - 4 \, d^{2}\right )} x^{3} + d^{2} + 4 \, {\left ({\left (c d - 2 \, d^{2}\right )} x^{4} + d^{2} x\right )} \sqrt {x^{4} - x} \sqrt {-\frac {c - d}{d}}}{c^{2} x^{6} - 2 \, c d x^{3} + d^{2}}\right ) - {\left ({\left (a - 2 \, b\right )} c - 2 \, a d\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x + 1\right )}{6 \, c^{2}}, \frac {2 \, \sqrt {x^{4} - x} a c x + 2 \, {\left (b c + a d\right )} \sqrt {\frac {c - d}{d}} \arctan \left (-\frac {2 \, \sqrt {x^{4} - x} d x \sqrt {\frac {c - d}{d}}}{{\left (c - 2 \, d\right )} x^{3} + d}\right ) - {\left ({\left (a - 2 \, b\right )} c - 2 \, a d\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x + 1\right )}{6 \, c^{2}}\right ] \]

[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 = 0.37 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {\left (b+a x^3\right ) \sqrt {-x+x^4}}{-d+c x^3} \, dx=\frac {\sqrt {x^{4} - x} a x}{3 \, c} - \frac {{\left (a c - 2 \, b c - 2 \, a d\right )} \log \left (\sqrt {-\frac {1}{x^{3}} + 1} + 1\right )}{6 \, c^{2}} + \frac {{\left (a c - 2 \, b c - 2 \, a d\right )} \log \left ({\left | \sqrt {-\frac {1}{x^{3}} + 1} - 1 \right |}\right )}{6 \, c^{2}} - \frac {2 \, {\left (b c^{2} + a c d - b c d - a d^{2}\right )} \arctan \left (\frac {d \sqrt {-\frac {1}{x^{3}} + 1}}{\sqrt {c d - d^{2}}}\right )}{3 \, \sqrt {c d - d^{2}} c^{2}} \]

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

[In]

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

[Out]

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