Integrand size = 30, antiderivative size = 221 \[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\frac {\left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b^2 d^3 n}-\frac {(5 b c+3 a d) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d n}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{8 b^{5/2} d^{7/2} n} \]
-1/8*(-a*d+b*c)*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(a+b*x^n)^(1 /2)/b^(1/2)/(c+d*x^n)^(1/2))/b^(5/2)/d^(7/2)/n-1/12*(3*a*d+5*b*c)*(a+b*x^n )^(3/2)*(c+d*x^n)^(1/2)/b^2/d^2/n+1/3*x^n*(a+b*x^n)^(3/2)*(c+d*x^n)^(1/2)/ b/d/n+1/8*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)*(a+b*x^n)^(1/2)*(c+d*x^n)^(1/2)/b^ 2/d^3/n
Time = 0.64 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.86 \[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\frac {b \sqrt {d} \sqrt {a+b x^n} \left (c+d x^n\right ) \left (-3 a^2 d^2+2 a b d \left (-2 c+d x^n\right )+b^2 \left (15 c^2-10 c d x^n+8 d^2 x^{2 n}\right )\right )-3 (b c-a d)^{3/2} \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {\frac {b \left (c+d x^n\right )}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b c-a d}}\right )}{24 b^3 d^{7/2} n \sqrt {c+d x^n}} \]
(b*Sqrt[d]*Sqrt[a + b*x^n]*(c + d*x^n)*(-3*a^2*d^2 + 2*a*b*d*(-2*c + d*x^n ) + b^2*(15*c^2 - 10*c*d*x^n + 8*d^2*x^(2*n))) - 3*(b*c - a*d)^(3/2)*(5*b^ 2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[(b*(c + d*x^n))/(b*c - a*d)]*ArcSinh[(Sq rt[d]*Sqrt[a + b*x^n])/Sqrt[b*c - a*d]])/(24*b^3*d^(7/2)*n*Sqrt[c + d*x^n] )
Time = 0.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {948, 101, 27, 90, 60, 66, 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.
\(\displaystyle \int \frac {x^{3 n-1} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\int \frac {x^{2 n} \sqrt {b x^n+a}}{\sqrt {d x^n+c}}dx^n}{n}\) |
\(\Big \downarrow \) 101 |
\(\displaystyle \frac {\frac {\int -\frac {\sqrt {b x^n+a} \left ((5 b c+3 a d) x^n+2 a c\right )}{2 \sqrt {d x^n+c}}dx^n}{3 b d}+\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d}}{n}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d}-\frac {\int \frac {\sqrt {b x^n+a} \left ((5 b c+3 a d) x^n+2 a c\right )}{\sqrt {d x^n+c}}dx^n}{6 b d}}{n}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d}-\frac {\frac {(3 a d+5 b c) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{2 b d}-\frac {3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \int \frac {\sqrt {b x^n+a}}{\sqrt {d x^n+c}}dx^n}{4 b d}}{6 b d}}{n}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d}-\frac {\frac {(3 a d+5 b c) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{2 b d}-\frac {3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{2 d}\right )}{4 b d}}{6 b d}}{n}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d}-\frac {\frac {(3 a d+5 b c) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{2 b d}-\frac {3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{d}-\frac {(b c-a d) \int \frac {1}{b-d x^{2 n}}d\frac {\sqrt {b x^n+a}}{\sqrt {d x^n+c}}}{d}\right )}{4 b d}}{6 b d}}{n}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {x^n \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{3 b d}-\frac {\frac {(3 a d+5 b c) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{2 b d}-\frac {3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 b d}}{6 b d}}{n}\) |
((x^n*(a + b*x^n)^(3/2)*Sqrt[c + d*x^n])/(3*b*d) - (((5*b*c + 3*a*d)*(a + b*x^n)^(3/2)*Sqrt[c + d*x^n])/(2*b*d) - (3*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^ 2)*((Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sq rt[a + b*x^n])/(Sqrt[b]*Sqrt[c + d*x^n])])/(Sqrt[b]*d^(3/2))))/(4*b*d))/(6 *b*d))/n
3.11.74.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
\[\int \frac {x^{-1+3 n} \sqrt {a +b \,x^{n}}}{\sqrt {c +d \,x^{n}}}d x\]
Time = 0.29 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.13 \[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2 \, n} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, \sqrt {b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2 \, n} + 15 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{96 \, b^{3} d^{4} n}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, \sqrt {-b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {-b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{2 \, {\left (b^{2} d^{2} x^{2 \, n} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2 \, n} + 15 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{48 \, b^{3} d^{4} n}\right ] \]
[-1/96*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*lo g(8*b^2*d^2*x^(2*n) + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*sqrt(b*d)*b*d*x ^n + (b*c + a*d)*sqrt(b*d))*sqrt(b*x^n + a)*sqrt(d*x^n + c) + 8*(b^2*c*d + a*b*d^2)*x^n) - 4*(8*b^3*d^3*x^(2*n) + 15*b^3*c^2*d - 4*a*b^2*c*d^2 - 3*a ^2*b*d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c ))/(b^3*d^4*n), 1/48*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3 )*sqrt(-b*d)*arctan(1/2*(2*sqrt(-b*d)*b*d*x^n + (b*c + a*d)*sqrt(-b*d))*sq rt(b*x^n + a)*sqrt(d*x^n + c)/(b^2*d^2*x^(2*n) + a*b*c*d + (b^2*c*d + a*b* d^2)*x^n)) + 2*(8*b^3*d^3*x^(2*n) + 15*b^3*c^2*d - 4*a*b^2*c*d^2 - 3*a^2*b *d^3 - 2*(5*b^3*c*d^2 - a*b^2*d^3)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/( b^3*d^4*n)]
\[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\int \frac {x^{3 n - 1} \sqrt {a + b x^{n}}}{\sqrt {c + d x^{n}}}\, dx \]
\[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\int { \frac {\sqrt {b x^{n} + a} x^{3 \, n - 1}}{\sqrt {d x^{n} + c}} \,d x } \]
\[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\int { \frac {\sqrt {b x^{n} + a} x^{3 \, n - 1}}{\sqrt {d x^{n} + c}} \,d x } \]
Timed out. \[ \int \frac {x^{-1+3 n} \sqrt {a+b x^n}}{\sqrt {c+d x^n}} \, dx=\int \frac {x^{3\,n-1}\,\sqrt {a+b\,x^n}}{\sqrt {c+d\,x^n}} \,d x \]