Integrand size = 24, antiderivative size = 123 \[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\frac {x^3 \sqrt {c+d x^6}}{6 b d}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x^3}{\sqrt {a} \sqrt {c+d x^6}}\right )}{3 b^2 \sqrt {b c-a d}}-\frac {(b c+2 a d) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{6 b^2 d^{3/2}} \]
-1/6*(2*a*d+b*c)*arctanh(x^3*d^(1/2)/(d*x^6+c)^(1/2))/b^2/d^(3/2)+1/3*a^(3 /2)*arctan(x^3*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^6+c)^(1/2))/b^2/(-a*d+b*c)^(1 /2)+1/6*x^3*(d*x^6+c)^(1/2)/b/d
Time = 1.41 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\frac {\frac {b x^3 \sqrt {c+d x^6}}{d}+\frac {2 a^{3/2} \arctan \left (\frac {a \sqrt {d}+b x^3 \left (\sqrt {d} x^3+\sqrt {c+d x^6}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{\sqrt {b c-a d}}-\frac {(b c+2 a d) \log \left (\sqrt {d} x^3+\sqrt {c+d x^6}\right )}{d^{3/2}}}{6 b^2} \]
((b*x^3*Sqrt[c + d*x^6])/d + (2*a^(3/2)*ArcTan[(a*Sqrt[d] + b*x^3*(Sqrt[d] *x^3 + Sqrt[c + d*x^6]))/(Sqrt[a]*Sqrt[b*c - a*d])])/Sqrt[b*c - a*d] - ((b *c + 2*a*d)*Log[Sqrt[d]*x^3 + Sqrt[c + d*x^6]])/d^(3/2))/(6*b^2)
Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {965, 381, 398, 224, 219, 291, 218}
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.
| \(\displaystyle \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx\) |
| \(\Big \downarrow \) 965 |
| \(\displaystyle \frac {1}{3} \int \frac {x^{12}}{\left (b x^6+a\right ) \sqrt {d x^6+c}}dx^3\) |
| \(\Big \downarrow \) 381 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^3 \sqrt {c+d x^6}}{2 b d}-\frac {\int \frac {(b c+2 a d) x^6+a c}{\left (b x^6+a\right ) \sqrt {d x^6+c}}dx^3}{2 b d}\right )\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^3 \sqrt {c+d x^6}}{2 b d}-\frac {\frac {(2 a d+b c) \int \frac {1}{\sqrt {d x^6+c}}dx^3}{b}-\frac {2 a^2 d \int \frac {1}{\left (b x^6+a\right ) \sqrt {d x^6+c}}dx^3}{b}}{2 b d}\right )\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^3 \sqrt {c+d x^6}}{2 b d}-\frac {\frac {(2 a d+b c) \int \frac {1}{1-d x^6}d\frac {x^3}{\sqrt {d x^6+c}}}{b}-\frac {2 a^2 d \int \frac {1}{\left (b x^6+a\right ) \sqrt {d x^6+c}}dx^3}{b}}{2 b d}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^3 \sqrt {c+d x^6}}{2 b d}-\frac {\frac {(2 a d+b c) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{b \sqrt {d}}-\frac {2 a^2 d \int \frac {1}{\left (b x^6+a\right ) \sqrt {d x^6+c}}dx^3}{b}}{2 b d}\right )\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^3 \sqrt {c+d x^6}}{2 b d}-\frac {\frac {(2 a d+b c) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{b \sqrt {d}}-\frac {2 a^2 d \int \frac {1}{a-(a d-b c) x^6}d\frac {x^3}{\sqrt {d x^6+c}}}{b}}{2 b d}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^3 \sqrt {c+d x^6}}{2 b d}-\frac {\frac {(2 a d+b c) \text {arctanh}\left (\frac {\sqrt {d} x^3}{\sqrt {c+d x^6}}\right )}{b \sqrt {d}}-\frac {2 a^{3/2} d \arctan \left (\frac {x^3 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^6}}\right )}{b \sqrt {b c-a d}}}{2 b d}\right )\) |
((x^3*Sqrt[c + d*x^6])/(2*b*d) - ((-2*a^(3/2)*d*ArcTan[(Sqrt[b*c - a*d]*x^ 3)/(Sqrt[a]*Sqrt[c + d*x^6])])/(b*Sqrt[b*c - a*d]) + ((b*c + 2*a*d)*ArcTan h[(Sqrt[d]*x^3)/Sqrt[c + d*x^6]])/(b*Sqrt[d]))/(2*b*d))/3
3.9.58.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q }, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 , p, q, x]
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]
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]
Time = 9.61 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.24
| method | result | size |
| pseudoelliptic | \(\frac {b \sqrt {\left (a d -b c \right ) a}\, \sqrt {d \,x^{6}+c}\, x^{3} \sqrt {d}+2 a^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{6}+c}\, a}{x^{3} \sqrt {\left (a d -b c \right ) a}}\right ) d^{\frac {3}{2}}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{6}+c}}{x^{3} \sqrt {d}}\right ) a d \sqrt {\left (a d -b c \right ) a}-\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{6}+c}}{x^{3} \sqrt {d}}\right ) b c \sqrt {\left (a d -b c \right ) a}}{6 b^{2} \sqrt {\left (a d -b c \right ) a}\, d^{\frac {3}{2}}}\) | \(152\) |
1/6*(b*((a*d-b*c)*a)^(1/2)*(d*x^6+c)^(1/2)*x^3*d^(1/2)+2*a^2*arctanh((d*x^ 6+c)^(1/2)/x^3*a/((a*d-b*c)*a)^(1/2))*d^(3/2)-2*arctanh((d*x^6+c)^(1/2)/x^ 3/d^(1/2))*a*d*((a*d-b*c)*a)^(1/2)-arctanh((d*x^6+c)^(1/2)/x^3/d^(1/2))*b* c*((a*d-b*c)*a)^(1/2))/b^2/((a*d-b*c)*a)^(1/2)/d^(3/2)
Time = 0.83 (sec) , antiderivative size = 739, normalized size of antiderivative = 6.01 \[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\left [\frac {2 \, \sqrt {d x^{6} + c} b d x^{3} + a d^{2} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{12} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{6} + a^{2} c^{2} + 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{9} - {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )} \sqrt {d x^{6} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{12} + 2 \, a b x^{6} + a^{2}}\right ) + {\left (b c + 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{6} + 2 \, \sqrt {d x^{6} + c} \sqrt {d} x^{3} - c\right )}{12 \, b^{2} d^{2}}, \frac {2 \, \sqrt {d x^{6} + c} b d x^{3} + a d^{2} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{12} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{6} + a^{2} c^{2} + 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{9} - {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )} \sqrt {d x^{6} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{12} + 2 \, a b x^{6} + a^{2}}\right ) + 2 \, {\left (b c + 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{3}}{\sqrt {d x^{6} + c}}\right )}{12 \, b^{2} d^{2}}, \frac {2 \, \sqrt {d x^{6} + c} b d x^{3} - 2 \, a d^{2} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{6} - a c\right )} \sqrt {d x^{6} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{9} + a c x^{3}\right )}}\right ) + {\left (b c + 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{6} + 2 \, \sqrt {d x^{6} + c} \sqrt {d} x^{3} - c\right )}{12 \, b^{2} d^{2}}, \frac {\sqrt {d x^{6} + c} b d x^{3} - a d^{2} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{6} - a c\right )} \sqrt {d x^{6} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{9} + a c x^{3}\right )}}\right ) + {\left (b c + 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{3}}{\sqrt {d x^{6} + c}}\right )}{6 \, b^{2} d^{2}}\right ] \]
[1/12*(2*sqrt(d*x^6 + c)*b*d*x^3 + a*d^2*sqrt(-a/(b*c - a*d))*log(((b^2*c^ 2 - 8*a*b*c*d + 8*a^2*d^2)*x^12 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^6 + a^2*c^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^9 - (a*b*c^2 - a^2*c*d)*x^3)*sqrt (d*x^6 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^12 + 2*a*b*x^6 + a^2)) + (b*c + 2 *a*d)*sqrt(d)*log(-2*d*x^6 + 2*sqrt(d*x^6 + c)*sqrt(d)*x^3 - c))/(b^2*d^2) , 1/12*(2*sqrt(d*x^6 + c)*b*d*x^3 + a*d^2*sqrt(-a/(b*c - a*d))*log(((b^2*c ^2 - 8*a*b*c*d + 8*a^2*d^2)*x^12 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^6 + a^2*c^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^9 - (a*b*c^2 - a^2*c*d)*x^3)*sqr t(d*x^6 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^12 + 2*a*b*x^6 + a^2)) + 2*(b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^3/sqrt(d*x^6 + c)))/(b^2*d^2), 1/12*(2 *sqrt(d*x^6 + c)*b*d*x^3 - 2*a*d^2*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^6 - a*c)*sqrt(d*x^6 + c)*sqrt(a/(b*c - a*d))/(a*d*x^9 + a*c*x^3) ) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^6 + 2*sqrt(d*x^6 + c)*sqrt(d)*x^3 - c ))/(b^2*d^2), 1/6*(sqrt(d*x^6 + c)*b*d*x^3 - a*d^2*sqrt(a/(b*c - a*d))*arc tan(-1/2*((b*c - 2*a*d)*x^6 - a*c)*sqrt(d*x^6 + c)*sqrt(a/(b*c - a*d))/(a* d*x^9 + a*c*x^3)) + (b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^3/sqrt(d*x^6 + c)))/(b^2*d^2)]
\[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\int \frac {x^{14}}{\left (a + b x^{6}\right ) \sqrt {c + d x^{6}}}\, dx \]
\[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\int { \frac {x^{14}}{{\left (b x^{6} + a\right )} \sqrt {d x^{6} + c}} \,d x } \]
Error detected during grading. Assigning place holder grade for now.
Time = 0.43 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.93 \[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\mathit {Recursive} \mathit {assumption} c \geq \frac {\sqrt {d x^{6} + c} x^{3}}{6 \, b d} - \frac {a^{2} \arctan \left (\frac {a \sqrt {d + \frac {c}{x^{6}}}}{\sqrt {a b c - a^{2} d}}\right )}{3 \, \sqrt {a b c - a^{2} d} b^{2} \mathrm {sgn}\left (x\right )} + \frac {{\left (2 \, a^{2} \sqrt {-d} d \arctan \left (\frac {a \sqrt {d}}{\sqrt {a b c - a^{2} d}}\right ) - \sqrt {a b c - a^{2} d} b c \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) - 2 \, \sqrt {a b c - a^{2} d} a d \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right )\right )} \mathrm {sgn}\left (x\right )}{6 \, \sqrt {a b c - a^{2} d} b^{2} \sqrt {-d} d} + \frac {{\left (b c + 2 \, a d\right )} \arctan \left (\frac {\sqrt {d + \frac {c}{x^{6}}}}{\sqrt {-d}}\right )}{6 \, b^{2} \sqrt {-d} d \mathrm {sgn}\left (x\right )} - \frac {d \mathit {ignored}}{t_{\mathit {nostep}}^{6}} \]
Recursive*assumption*c >= 1/6*sqrt(d*x^6 + c)*x^3/(b*d) - 1/3*a^2*arctan(a *sqrt(d + c/x^6)/sqrt(a*b*c - a^2*d))/(sqrt(a*b*c - a^2*d)*b^2*sgn(x)) + 1 /6*(2*a^2*sqrt(-d)*d*arctan(a*sqrt(d)/sqrt(a*b*c - a^2*d)) - sqrt(a*b*c - a^2*d)*b*c*arctan(sqrt(d)/sqrt(-d)) - 2*sqrt(a*b*c - a^2*d)*a*d*arctan(sqr t(d)/sqrt(-d)))*sgn(x)/(sqrt(a*b*c - a^2*d)*b^2*sqrt(-d)*d) + 1/6*(b*c + 2 *a*d)*arctan(sqrt(d + c/x^6)/sqrt(-d))/(b^2*sqrt(-d)*d*sgn(x)) - d*ignored /t_nostep^6
Timed out. \[ \int \frac {x^{14}}{\left (a+b x^6\right ) \sqrt {c+d x^6}} \, dx=\int \frac {x^{14}}{\left (b\,x^6+a\right )\,\sqrt {d\,x^6+c}} \,d x \]