\(\int \frac {(a x+b x^3)^{3/2}}{x^8} \, dx\) [56]

Optimal result

Integrand size = 17, antiderivative size = 163 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {4 b^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{77 a^{5/4} \sqrt {a x+b x^3}} \]

[Out]

-2/11*(b*x^3+a*x)^(3/2)/x^7-12/77*b*(b*x^3+a*x)^(1/2)/x^4-8/77*b^2*(b*x^3+a*x)^(1/2)/a/x^2-4/77*b^(11/4)*(cos(
2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1
/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(5/4)
/(b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2045, 2050, 2036, 335, 226} \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {4 b^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{77 a^{5/4} \sqrt {a x+b x^3}}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {12 b \sqrt {a x+b x^3}}{77 x^4} \]

[In]

Int[(a*x + b*x^3)^(3/2)/x^8,x]

[Out]

(-12*b*Sqrt[a*x + b*x^3])/(77*x^4) - (8*b^2*Sqrt[a*x + b*x^3])/(77*a*x^2) - (2*(a*x + b*x^3)^(3/2))/(11*x^7) -
(4*b^(11/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/
4)*Sqrt[x])/a^(1/4)], 1/2])/(77*a^(5/4)*Sqrt[a*x + b*x^3])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(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] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2045

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

Rule 2050

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] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}+\frac {1}{11} (6 b) \int \frac {\sqrt {a x+b x^3}}{x^5} \, dx \\ & = -\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}+\frac {1}{77} \left (12 b^2\right ) \int \frac {1}{x^2 \sqrt {a x+b x^3}} \, dx \\ & = -\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {\left (4 b^3\right ) \int \frac {1}{\sqrt {a x+b x^3}} \, dx}{77 a} \\ & = -\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {\left (4 b^3 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{77 a \sqrt {a x+b x^3}} \\ & = -\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {\left (8 b^3 \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{77 a \sqrt {a x+b x^3}} \\ & = -\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {4 b^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{77 a^{5/4} \sqrt {a x+b x^3}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {2 a \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {3}{2},-\frac {7}{4},-\frac {b x^2}{a}\right )}{11 x^6 \sqrt {1+\frac {b x^2}{a}}} \]

[In]

Integrate[(a*x + b*x^3)^(3/2)/x^8,x]

[Out]

(-2*a*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-11/4, -3/2, -7/4, -((b*x^2)/a)])/(11*x^6*Sqrt[1 + (b*x^2)/a])

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {2 \left (b \,x^{2}+a \right ) \left (4 b^{2} x^{4}+13 a b \,x^{2}+7 a^{2}\right )}{77 x^{5} \sqrt {x \left (b \,x^{2}+a \right )}\, a}-\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a \sqrt {b \,x^{3}+a x}}\)	\(160\)
default	\(-\frac {2 a \sqrt {b \,x^{3}+a x}}{11 x^{6}}-\frac {26 b \sqrt {b \,x^{3}+a x}}{77 x^{4}}-\frac {8 b^{2} \sqrt {b \,x^{3}+a x}}{77 a \,x^{2}}-\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a \sqrt {b \,x^{3}+a x}}\)	\(169\)
elliptic	\(-\frac {2 a \sqrt {b \,x^{3}+a x}}{11 x^{6}}-\frac {26 b \sqrt {b \,x^{3}+a x}}{77 x^{4}}-\frac {8 b^{2} \sqrt {b \,x^{3}+a x}}{77 a \,x^{2}}-\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a \sqrt {b \,x^{3}+a x}}\)	\(169\)

[In]

int((b*x^3+a*x)^(3/2)/x^8,x,method=_RETURNVERBOSE)

[Out]

-2/77*(b*x^2+a)*(4*b^2*x^4+13*a*b*x^2+7*a^2)/x^5/(x*(b*x^2+a))^(1/2)/a-4/77*b^2/a*(-a*b)^(1/2)*((x+(-a*b)^(1/2
)/b)/(-a*b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b*x^3+a*x)^
(1/2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.36 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {2 \, {\left (4 \, b^{\frac {5}{2}} x^{6} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (4 \, b^{2} x^{4} + 13 \, a b x^{2} + 7 \, a^{2}\right )} \sqrt {b x^{3} + a x}\right )}}{77 \, a x^{6}} \]

[In]

integrate((b*x^3+a*x)^(3/2)/x^8,x, algorithm="fricas")

[Out]

-2/77*(4*b^(5/2)*x^6*weierstrassPInverse(-4*a/b, 0, x) + (4*b^2*x^4 + 13*a*b*x^2 + 7*a^2)*sqrt(b*x^3 + a*x))/(
a*x^6)

Sympy [F]

\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \]

[In]

integrate((b*x**3+a*x)**(3/2)/x**8,x)

[Out]

Integral((x*(a + b*x**2))**(3/2)/x**8, x)

Maxima [F]

\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \]

[In]

integrate((b*x^3+a*x)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^8, x)

Giac [F]

\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \]

[In]

integrate((b*x^3+a*x)^(3/2)/x^8,x, algorithm="giac")

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^8, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x\right )}^{3/2}}{x^8} \,d x \]

[In]

int((a*x + b*x^3)^(3/2)/x^8,x)

[Out]

int((a*x + b*x^3)^(3/2)/x^8, x)