\(\int \frac {1}{(a x^2+b x^3)^{7/4}} \, dx\) [118]

Optimal result

Integrand size = 15, antiderivative size = 174 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}-\frac {26 \sqrt [4]{a x^2+b x^3}}{15 a^2 x^3}+\frac {13 b \sqrt [4]{a x^2+b x^3}}{5 a^3 x^2}-\frac {13 b^2 \sqrt [4]{a x^2+b x^3}}{2 a^4 x}-\frac {13 b^{5/2} x^{3/2} \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),2\right )}{2 a^{7/2} \left (a x^2+b x^3\right )^{3/4}} \] Output:

4/3/a/x/(b*x^3+a*x^2)^(3/4)-26/15*(b*x^3+a*x^2)^(1/4)/a^2/x^3+13/5*b*(b*x^ 
3+a*x^2)^(1/4)/a^3/x^2-13/2*b^2*(b*x^3+a*x^2)^(1/4)/a^4/x-13/2*b^(5/2)*x^( 
3/2)*(1+b*x/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x^(1/2)/a^(1/2)),2 
^(1/2))/a^(7/2)/(b*x^3+a*x^2)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=-\frac {2 \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {7}{4},-\frac {3}{2},-\frac {b x}{a}\right )}{5 a x \left (x^2 (a+b x)\right )^{3/4}} \] Input:

Integrate[(a*x^2 + b*x^3)^(-7/4),x]
 

Output:

(-2*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[-5/2, 7/4, -3/2, -((b*x)/a)])/(5 
*a*x*(x^2*(a + b*x))^(3/4))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1912, 1931, 1931, 1917, 61, 73, 765, 762}

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.

\(\displaystyle \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx\)

\(\Big \downarrow \) 1912

\(\displaystyle \frac {13 \int \frac {1}{x^2 \left (b x^3+a x^2\right )^{3/4}}dx}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {13 \left (-\frac {9 b \int \frac {1}{x \left (b x^3+a x^2\right )^{3/4}}dx}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b \int \frac {1}{\left (b x^3+a x^2\right )^{3/4}}dx}{6 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\right )}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b x^{3/2} (a+b x)^{3/4} \int \frac {1}{x^{3/2} (a+b x)^{3/4}}dx}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\right )}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)^{3/4}}dx}{2 a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\right )}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \int \frac {1}{\sqrt {\frac {a+b x}{b}-\frac {a}{b}}}d\sqrt [4]{a+b x}}{a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\right )}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \int \frac {1}{\sqrt {1-\frac {a+b x}{a}}}d\sqrt [4]{a+b x}}{a \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\right )}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{a+b x}}{\sqrt [4]{a}}\right ),-1\right )}{a^{3/4} \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\right )}{10 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{5 a x^3}\right )}{3 a}+\frac {4}{3 a x \left (a x^2+b x^3\right )^{3/4}}\)

Input:

Int[(a*x^2 + b*x^3)^(-7/4),x]
 

Output:

4/(3*a*x*(a*x^2 + b*x^3)^(3/4)) + (13*((-2*(a*x^2 + b*x^3)^(1/4))/(5*a*x^3 
) - (9*b*((-2*(a*x^2 + b*x^3)^(1/4))/(3*a*x^2) - (5*b*x^(3/2)*(a + b*x)^(3 
/4)*((-2*(a + b*x)^(1/4))/(a*Sqrt[x]) - (2*Sqrt[1 - (a + b*x)/a]*EllipticF 
[ArcSin[(a + b*x)^(1/4)/a^(1/4)], -1])/(a^(3/4)*Sqrt[-(a/b) + (a + b*x)/b] 
)))/(6*a*(a*x^2 + b*x^3)^(3/4))))/(10*a)))/(3*a)
 

Defintions of rubi rules used

rule 61

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]
 

rule 73

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]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

rule 1912

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[-(a*x^j + 
 b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*x^(j - 1)), x] + Simp[(n*p + n - j + 1)/ 
(a*(n - j)*(p + 1))   Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[{a, 
b}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1931

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] 
 &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ 
m + j*p + 1, 0]
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {7}{4}}}d x\]

Input:

int(1/(b*x^3+a*x^2)^(7/4),x)
 

Output:

int(1/(b*x^3+a*x^2)^(7/4),x)
 

Fricas [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(7/4),x, algorithm="fricas")
 

Output:

integral((b*x^3 + a*x^2)^(1/4)/(b^2*x^6 + 2*a*b*x^5 + a^2*x^4), x)
 

Sympy [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {7}{4}}}\, dx \] Input:

integrate(1/(b*x**3+a*x**2)**(7/4),x)
 

Output:

Integral((a*x**2 + b*x**3)**(-7/4), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(7/4),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(-7/4), x)
 

Giac [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(7/4),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a*x^2)^(-7/4), x)
 

Mupad [B] (verification not implemented)

Time = 10.55 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=-\frac {2\,x\,{\left (\frac {b\,x}{a}+1\right )}^{7/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {7}{4};\ -\frac {3}{2};\ -\frac {b\,x}{a}\right )}{5\,{\left (b\,x^3+a\,x^2\right )}^{7/4}} \] Input:

int(1/(a*x^2 + b*x^3)^(7/4),x)
 

Output:

-(2*x*((b*x)/a + 1)^(7/4)*hypergeom([-5/2, 7/4], -3/2, -(b*x)/a))/(5*(a*x^ 
2 + b*x^3)^(7/4))
 

Reduce [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{7/4}} \, dx=\int \frac {\sqrt {x}\, \left (b x +a \right )^{\frac {3}{4}}}{\sqrt {b x +a}\, a^{2} x^{4}+2 \sqrt {b x +a}\, a b \,x^{5}+\sqrt {b x +a}\, b^{2} x^{6}}d x \] Input:

int(1/(b*x^3+a*x^2)^(7/4),x)
 

Output:

int((sqrt(x)*(a + b*x)**(3/4))/(sqrt(a + b*x)*a**2*x**4 + 2*sqrt(a + b*x)* 
a*b*x**5 + sqrt(a + b*x)*b**2*x**6),x)