Integrand size = 17, antiderivative size = 116 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{a x+b x^2}}{7 a x^2}+\frac {8 b \sqrt [4]{a x+b x^2}}{7 a^2 x}-\frac {16 b \left (\frac {b x}{a+b x}\right )^{3/4} (a+b x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right ),2\right )}{7 a^{5/2} \left (a x+b x^2\right )^{3/4}} \] Output:
-4/7*(b*x^2+a*x)^(1/4)/a/x^2+8/7*b*(b*x^2+a*x)^(1/4)/a^2/x-16/7*b*(b*x/(b* x+a))^(3/4)*(b*x+a)^(3/2)*InverseJacobiAM(1/2*arcsin(a^(1/2)/(b*x+a)^(1/2) ),2^(1/2))/a^(5/2)/(b*x^2+a*x)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=-\frac {4 \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{4},-\frac {3}{4},-\frac {b x}{a}\right )}{7 x (x (a+b x))^{3/4}} \] Input:
Integrate[1/(x^2*(a*x + b*x^2)^(3/4)),x]
Output:
(-4*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[-7/4, 3/4, -3/4, -((b*x)/a)])/(7 *x*(x*(a + b*x))^(3/4))
Time = 0.47 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.25, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1137, 61, 61, 73, 768, 858, 807, 229}
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.
\(\displaystyle \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 1137 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \int \frac {1}{x^{11/4} (a+b x)^{3/4}}dx}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \int \frac {1}{x^{7/4} (a+b x)^{3/4}}dx}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \left (-\frac {2 b \int \frac {1}{x^{3/4} (a+b x)^{3/4}}dx}{3 a}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \left (-\frac {8 b \int \frac {1}{(a+b x)^{3/4}}d\sqrt [4]{x}}{3 a}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \left (-\frac {8 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{3/4} x^{3/4}}d\sqrt [4]{x}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \left (\frac {8 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{x}}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \left (\frac {4 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{3/4}}d\sqrt {x}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {x^{3/4} (a+b x)^{3/4} \left (-\frac {6 b \left (\frac {8 b^{3/2} x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{\left (a x+b x^2\right )^{3/4}}\) |
Input:
Int[1/(x^2*(a*x + b*x^2)^(3/4)),x]
Output:
(x^(3/4)*(a + b*x)^(3/4)*((-4*(a + b*x)^(1/4))/(7*a*x^(7/4)) - (6*b*((-4*( a + b*x)^(1/4))/(3*a*x^(3/4)) + (8*b^(3/2)*(1 + a/(b*x))^(3/4)*x^(3/4)*Ell ipticF[ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]/2, 2])/(3*a^(3/2)*(a + b*x)^(3/4) )))/(7*a)))/(a*x + b*x^2)^(3/4)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 /4)/(a + b*x^4)^(3/4)) Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ [{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p)) Int[x^(m + p)*(b + c*x)^ p, x], x] /; FreeQ[{b, c, e, m}, x]
\[\int \frac {1}{x^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{4}}}d x\]
Input:
int(1/x^2/(b*x^2+a*x)^(3/4),x)
Output:
int(1/x^2/(b*x^2+a*x)^(3/4),x)
\[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {3}{4}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a*x)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a*x)^(1/4)/(b*x^4 + a*x^3), x)
\[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=\int \frac {1}{x^{2} \left (x \left (a + b x\right )\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(1/x**2/(b*x**2+a*x)**(3/4),x)
Output:
Integral(1/(x**2*(x*(a + b*x))**(3/4)), x)
\[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {3}{4}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a*x)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a*x)^(3/4)*x^2), x)
\[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {3}{4}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a*x)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a*x)^(3/4)*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=\int \frac {1}{x^2\,{\left (b\,x^2+a\,x\right )}^{3/4}} \,d x \] Input:
int(1/(x^2*(a*x + b*x^2)^(3/4)),x)
Output:
int(1/(x^2*(a*x + b*x^2)^(3/4)), x)
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{3/4}} \, dx=\frac {4 \left (b x +a \right )^{\frac {1}{4}} \left (4 b x -a \right )}{5 x^{\frac {5}{4}} \sqrt {x}\, a^{2}} \] Input:
int(1/x^2/(b*x^2+a*x)^(3/4),x)
Output:
(4*x**(3/4)*(a + b*x)**(1/4)*( - a + 4*b*x))/(5*sqrt(x)*a**2*x**2)