Integrand size = 13, antiderivative size = 154 \[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=\frac {4}{5 b \left (b x+c x^2\right )^{5/4}}+\frac {96 c}{5 b^3 \sqrt [4]{b x+c x^2}}+\frac {8}{b^2 x \sqrt [4]{b x+c x^2}}-\frac {48 \left (b x+c x^2\right )^{3/4}}{5 b^3 x^2}-\frac {96 c \sqrt [4]{\frac {c x}{b+c x}} \sqrt {b+c x} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b}}{\sqrt {b+c x}}\right )\right |2\right )}{5 b^{7/2} \sqrt [4]{b x+c x^2}} \] Output:
4/5/b/(c*x^2+b*x)^(5/4)+96/5*c/b^3/(c*x^2+b*x)^(1/4)+8/b^2/x/(c*x^2+b*x)^( 1/4)-48/5*(c*x^2+b*x)^(3/4)/b^3/x^2-96/5*c*(c*x/(c*x+b))^(1/4)*(c*x+b)^(1/ 2)*EllipticE(sin(1/2*arcsin(b^(1/2)/(c*x+b)^(1/2))),2^(1/2))/b^(7/2)/(c*x^ 2+b*x)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.32 \[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=-\frac {4 \sqrt [4]{1+\frac {c x}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {9}{4},-\frac {1}{4},-\frac {c x}{b}\right )}{5 b^2 x \sqrt [4]{x (b+c x)}} \] Input:
Integrate[(b*x + c*x^2)^(-9/4),x]
Output:
(-4*(1 + (c*x)/b)^(1/4)*Hypergeometric2F1[-5/4, 9/4, -1/4, -((c*x)/b)])/(5 *b^2*x*(x*(b + c*x))^(1/4))
Time = 0.46 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1089, 1089, 1093, 1090, 226}
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.
\(\displaystyle \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle -\frac {12 c \int \frac {1}{\left (c x^2+b x\right )^{5/4}}dx}{5 b^2}-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle -\frac {12 c \left (\frac {4 c \int \frac {1}{\sqrt [4]{c x^2+b x}}dx}{b^2}-\frac {4 (b+2 c x)}{b^2 \sqrt [4]{b x+c x^2}}\right )}{5 b^2}-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 1093 |
\(\displaystyle -\frac {12 c \left (\frac {4 c \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} \int \frac {1}{\sqrt [4]{-\frac {c^2 x^2}{b^2}-\frac {c x}{b}}}dx}{b^2 \sqrt [4]{b x+c x^2}}-\frac {4 (b+2 c x)}{b^2 \sqrt [4]{b x+c x^2}}\right )}{5 b^2}-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle -\frac {12 c \left (-\frac {2 \sqrt {2} \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} \int \frac {1}{\sqrt [4]{1-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}}}d\left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )}{c \sqrt [4]{b x+c x^2}}-\frac {4 (b+2 c x)}{b^2 \sqrt [4]{b x+c x^2}}\right )}{5 b^2}-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 226 |
\(\displaystyle -\frac {12 c \left (-\frac {4 \sqrt {2} \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )}{c}\right )\right |2\right )}{b \sqrt [4]{b x+c x^2}}-\frac {4 (b+2 c x)}{b^2 \sqrt [4]{b x+c x^2}}\right )}{5 b^2}-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}\) |
Input:
Int[(b*x + c*x^2)^(-9/4),x]
Output:
(-4*(b + 2*c*x))/(5*b^2*(b*x + c*x^2)^(5/4)) - (12*c*((-4*(b + 2*c*x))/(b^ 2*(b*x + c*x^2)^(1/4)) - (4*Sqrt[2]*(-((c*(b*x + c*x^2))/b^2))^(1/4)*Ellip ticE[ArcSin[(b*(-(c/b) - (2*c^2*x)/b^2))/c]/2, 2])/(b*(b*x + c*x^2)^(1/4)) ))/(5*b^2)
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] ))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b*x + c*x^2)^p/((- c)*((b*x + c*x^2)/b^2))^p Int[((-c)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; F reeQ[{b, c}, x] && (IntegerQ[4*p] || IntegerQ[3*p])
\[\int \frac {1}{\left (c \,x^{2}+b x \right )^{\frac {9}{4}}}d x\]
Input:
int(1/(c*x^2+b*x)^(9/4),x)
Output:
int(1/(c*x^2+b*x)^(9/4),x)
\[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate(1/(c*x^2+b*x)^(9/4),x, algorithm="fricas")
Output:
integral((c*x^2 + b*x)^(3/4)/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*x^ 3), x)
\[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=\int \frac {1}{\left (b x + c x^{2}\right )^{\frac {9}{4}}}\, dx \] Input:
integrate(1/(c*x**2+b*x)**(9/4),x)
Output:
Integral((b*x + c*x**2)**(-9/4), x)
\[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate(1/(c*x^2+b*x)^(9/4),x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x)^(-9/4), x)
\[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate(1/(c*x^2+b*x)^(9/4),x, algorithm="giac")
Output:
integrate((c*x^2 + b*x)^(-9/4), x)
Time = 9.64 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.23 \[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=-\frac {4\,x\,{\left (\frac {c\,x}{b}+1\right )}^{9/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {9}{4};\ -\frac {1}{4};\ -\frac {c\,x}{b}\right )}{5\,{\left (c\,x^2+b\,x\right )}^{9/4}} \] Input:
int(1/(b*x + c*x^2)^(9/4),x)
Output:
-(4*x*((c*x)/b + 1)^(9/4)*hypergeom([-5/4, 9/4], -1/4, -(c*x)/b))/(5*(b*x + c*x^2)^(9/4))
Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx=\frac {-\frac {128}{15} c^{2} x^{2}-\frac {32}{3} b c x -\frac {4}{3} b^{2}}{x^{\frac {3}{4}} \left (c x +b \right )^{\frac {3}{4}} \sqrt {x}\, \sqrt {c x +b}\, b^{3}} \] Input:
int(1/(c*x^2+b*x)^(9/4),x)
Output:
(4*x**(1/4)*(b + c*x)**(1/4)*( - 5*b**2 - 40*b*c*x - 32*c**2*x**2))/(15*sq rt(x)*sqrt(b + c*x)*b**3*x*(b + c*x))