Integrand size = 22, antiderivative size = 153 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {(b c-a d) x^{13}}{13 a b \left (a+b x^4\right )^{13/4}}-\frac {d x^9}{9 b^2 \left (a+b x^4\right )^{9/4}}-\frac {d x^5}{5 b^3 \left (a+b x^4\right )^{5/4}}-\frac {d x}{b^4 \sqrt [4]{a+b x^4}}+\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{17/4}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{17/4}} \] Output:
1/13*(-a*d+b*c)*x^13/a/b/(b*x^4+a)^(13/4)-1/9*d*x^9/b^2/(b*x^4+a)^(9/4)-1/ 5*d*x^5/b^3/(b*x^4+a)^(5/4)-d*x/b^4/(b*x^4+a)^(1/4)+1/2*d*arctan(b^(1/4)*x /(b*x^4+a)^(1/4))/b^(17/4)+1/2*d*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(17/ 4)
Time = 6.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.81 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {-\frac {2 \sqrt [4]{b} x \left (585 a^4 d+1872 a^3 b d x^4+2054 a^2 b^2 d x^8-45 b^4 c x^{12}+812 a b^3 d x^{12}\right )}{a \left (a+b x^4\right )^{13/4}}+585 d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+585 d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{1170 b^{17/4}} \] Input:
Integrate[(x^12*(c + d*x^4))/(a + b*x^4)^(17/4),x]
Output:
((-2*b^(1/4)*x*(585*a^4*d + 1872*a^3*b*d*x^4 + 2054*a^2*b^2*d*x^8 - 45*b^4 *c*x^12 + 812*a*b^3*d*x^12))/(a*(a + b*x^4)^(13/4)) + 585*d*ArcTan[(b^(1/4 )*x)/(a + b*x^4)^(1/4)] + 585*d*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(1 170*b^(17/4))
Time = 0.53 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {954, 817, 817, 817, 770, 756, 216, 219}
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.
\(\displaystyle \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx\) |
\(\Big \downarrow \) 954 |
\(\displaystyle \frac {d \int \frac {x^{12}}{\left (b x^4+a\right )^{13/4}}dx}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {d \left (\frac {\int \frac {x^8}{\left (b x^4+a\right )^{9/4}}dx}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {d \left (\frac {\frac {\int \frac {x^4}{\left (b x^4+a\right )^{5/4}}dx}{b}-\frac {x^5}{5 b \left (a+b x^4\right )^{5/4}}}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {d \left (\frac {\frac {\frac {\int \frac {1}{\sqrt [4]{b x^4+a}}dx}{b}-\frac {x}{b \sqrt [4]{a+b x^4}}}{b}-\frac {x^5}{5 b \left (a+b x^4\right )^{5/4}}}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {d \left (\frac {\frac {\frac {\int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{b}-\frac {x}{b \sqrt [4]{a+b x^4}}}{b}-\frac {x^5}{5 b \left (a+b x^4\right )^{5/4}}}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {d \left (\frac {\frac {\frac {\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{b}-\frac {x}{b \sqrt [4]{a+b x^4}}}{b}-\frac {x^5}{5 b \left (a+b x^4\right )^{5/4}}}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {d \left (\frac {\frac {\frac {\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}}{b}-\frac {x}{b \sqrt [4]{a+b x^4}}}{b}-\frac {x^5}{5 b \left (a+b x^4\right )^{5/4}}}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {d \left (\frac {\frac {\frac {\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}}{b}-\frac {x}{b \sqrt [4]{a+b x^4}}}{b}-\frac {x^5}{5 b \left (a+b x^4\right )^{5/4}}}{b}-\frac {x^9}{9 b \left (a+b x^4\right )^{9/4}}\right )}{b}+\frac {x^{13} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
Input:
Int[(x^12*(c + d*x^4))/(a + b*x^4)^(17/4),x]
Output:
((b*c - a*d)*x^13)/(13*a*b*(a + b*x^4)^(13/4)) + (d*(-1/9*x^9/(b*(a + b*x^ 4)^(9/4)) + (-1/5*x^5/(b*(a + b*x^4)^(5/4)) + (-(x/(b*(a + b*x^4)^(1/4))) + (ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x) /(a + b*x^4)^(1/4)]/(2*b^(1/4)))/b)/b)/b))/b
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(b*c - a*d)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b* e*(m + 1))), x] + Simp[d/b Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; Fre eQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && NeQ[m, -1]
Time = 0.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.89
method | result | size |
pseudoelliptic | \(\frac {-\frac {3248 a \,b^{\frac {13}{4}} d \,x^{13}}{585}+\frac {4 b^{\frac {17}{4}} c \,x^{13}}{13}+d \left (-\frac {64 a^{2} b^{\frac {5}{4}} x^{5}}{5}-\frac {632 a \,b^{\frac {9}{4}} x^{9}}{45}-4 a^{3} x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {13}{4}} \left (\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right )\right )\right ) a}{4 b^{\frac {17}{4}} \left (b \,x^{4}+a \right )^{\frac {13}{4}} a}\) | \(136\) |
Input:
int(x^12*(d*x^4+c)/(b*x^4+a)^(17/4),x,method=_RETURNVERBOSE)
Output:
1/4*(-3248/585*a*b^(13/4)*d*x^13+4/13*b^(17/4)*c*x^13+d*(-64/5*a^2*b^(5/4) *x^5-632/45*a*b^(9/4)*x^9-4*a^3*x*b^(1/4)+(b*x^4+a)^(13/4)*(ln((x*b^(1/4)+ (b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))-2*arctan((b*x^4+a)^(1/4)/x/ b^(1/4))))*a)/b^(17/4)/(b*x^4+a)^(13/4)/a
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 494, normalized size of antiderivative = 3.23 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {585 \, {\left (a b^{8} x^{16} + 4 \, a^{2} b^{7} x^{12} + 6 \, a^{3} b^{6} x^{8} + 4 \, a^{4} b^{5} x^{4} + a^{5} b^{4}\right )} \left (\frac {d^{4}}{b^{17}}\right )^{\frac {1}{4}} \log \left (\frac {b^{13} x \left (\frac {d^{4}}{b^{17}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 585 \, {\left (a b^{8} x^{16} + 4 \, a^{2} b^{7} x^{12} + 6 \, a^{3} b^{6} x^{8} + 4 \, a^{4} b^{5} x^{4} + a^{5} b^{4}\right )} \left (\frac {d^{4}}{b^{17}}\right )^{\frac {1}{4}} \log \left (-\frac {b^{13} x \left (\frac {d^{4}}{b^{17}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 585 \, {\left (i \, a b^{8} x^{16} + 4 i \, a^{2} b^{7} x^{12} + 6 i \, a^{3} b^{6} x^{8} + 4 i \, a^{4} b^{5} x^{4} + i \, a^{5} b^{4}\right )} \left (\frac {d^{4}}{b^{17}}\right )^{\frac {1}{4}} \log \left (\frac {i \, b^{13} x \left (\frac {d^{4}}{b^{17}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 585 \, {\left (-i \, a b^{8} x^{16} - 4 i \, a^{2} b^{7} x^{12} - 6 i \, a^{3} b^{6} x^{8} - 4 i \, a^{4} b^{5} x^{4} - i \, a^{5} b^{4}\right )} \left (\frac {d^{4}}{b^{17}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, b^{13} x \left (\frac {d^{4}}{b^{17}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 4 \, {\left (2054 \, a^{2} b^{2} d x^{9} - {\left (45 \, b^{4} c - 812 \, a b^{3} d\right )} x^{13} + 1872 \, a^{3} b d x^{5} + 585 \, a^{4} d x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{2340 \, {\left (a b^{8} x^{16} + 4 \, a^{2} b^{7} x^{12} + 6 \, a^{3} b^{6} x^{8} + 4 \, a^{4} b^{5} x^{4} + a^{5} b^{4}\right )}} \] Input:
integrate(x^12*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="fricas")
Output:
1/2340*(585*(a*b^8*x^16 + 4*a^2*b^7*x^12 + 6*a^3*b^6*x^8 + 4*a^4*b^5*x^4 + a^5*b^4)*(d^4/b^17)^(1/4)*log((b^13*x*(d^4/b^17)^(3/4) + (b*x^4 + a)^(1/4 )*d^3)/x) - 585*(a*b^8*x^16 + 4*a^2*b^7*x^12 + 6*a^3*b^6*x^8 + 4*a^4*b^5*x ^4 + a^5*b^4)*(d^4/b^17)^(1/4)*log(-(b^13*x*(d^4/b^17)^(3/4) - (b*x^4 + a) ^(1/4)*d^3)/x) - 585*(I*a*b^8*x^16 + 4*I*a^2*b^7*x^12 + 6*I*a^3*b^6*x^8 + 4*I*a^4*b^5*x^4 + I*a^5*b^4)*(d^4/b^17)^(1/4)*log((I*b^13*x*(d^4/b^17)^(3/ 4) + (b*x^4 + a)^(1/4)*d^3)/x) - 585*(-I*a*b^8*x^16 - 4*I*a^2*b^7*x^12 - 6 *I*a^3*b^6*x^8 - 4*I*a^4*b^5*x^4 - I*a^5*b^4)*(d^4/b^17)^(1/4)*log((-I*b^1 3*x*(d^4/b^17)^(3/4) + (b*x^4 + a)^(1/4)*d^3)/x) - 4*(2054*a^2*b^2*d*x^9 - (45*b^4*c - 812*a*b^3*d)*x^13 + 1872*a^3*b*d*x^5 + 585*a^4*d*x)*(b*x^4 + a)^(3/4))/(a*b^8*x^16 + 4*a^2*b^7*x^12 + 6*a^3*b^6*x^8 + 4*a^4*b^5*x^4 + a ^5*b^4)
Timed out. \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:
integrate(x**12*(d*x**4+c)/(b*x**4+a)**(17/4),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.06 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {c x^{13}}{13 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a} - \frac {1}{2340} \, {\left (\frac {4 \, {\left (45 \, b^{3} + \frac {65 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{2} b}{x^{8}} + \frac {585 \, {\left (b x^{4} + a\right )}^{3}}{x^{12}}\right )} x^{13}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} b^{4}} + \frac {585 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{4}}\right )} d \] Input:
integrate(x^12*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="maxima")
Output:
1/13*c*x^13/((b*x^4 + a)^(13/4)*a) - 1/2340*(4*(45*b^3 + 65*(b*x^4 + a)*b^ 2/x^4 + 117*(b*x^4 + a)^2*b/x^8 + 585*(b*x^4 + a)^3/x^12)*x^13/((b*x^4 + a )^(13/4)*b^4) + 585*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log (-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1/4) )/b^4)*d
\[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{12}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:
integrate(x^12*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^12/(b*x^4 + a)^(17/4), x)
Timed out. \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int \frac {x^{12}\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{17/4}} \,d x \] Input:
int((x^12*(c + d*x^4))/(a + b*x^4)^(17/4),x)
Output:
int((x^12*(c + d*x^4))/(a + b*x^4)^(17/4), x)
\[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{16}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {x^{12}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) c \] Input:
int(x^12*(d*x^4+c)/(b*x^4+a)^(17/4),x)
Output:
int(x**16/((a + b*x**4)**(1/4)*a**4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x**8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16),x)*d + int(x**12/((a + b*x**4)**(1/4)*a* *4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x **8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16) ,x)*c