\(\int \frac {(a x+b x^3)^{3/2}}{x^7} \, dx\) [55]

Optimal result

Integrand size = 17, antiderivative size = 306 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^7} \, dx=\frac {8 b^{5/2} x \left (a+b x^2\right )}{15 a \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}-\frac {8 b^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a x+b x^3}}+\frac {4 b^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a x+b x^3}} \]

[Out]

-2/9*(b*x^3+a*x)^(3/2)/x^6+8/15*b^(5/2)*x*(b*x^2+a)/a/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)-4/15*b*(b*x^3+a*x)
^(1/2)/x^3-8/15*b^2*(b*x^3+a*x)^(1/2)/a/x-8/15*b^(9/4)*(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)))*EllipticE(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^(3/4)/(b*x^3+a*x)^(1/2)+4/15*b^(9/4)*(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^(3/4)/(b*x^3+a*x)^
(1/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2045, 2050, 2057, 335, 311, 226, 1210} \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^7} \, dx=\frac {4 b^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a x+b x^3}}-\frac {8 b^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a x+b x^3}}+\frac {8 b^{5/2} x \left (a+b x^2\right )}{15 a \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6} \]

[In]

Int[(a*x + b*x^3)^(3/2)/x^7,x]

[Out]

(8*b^(5/2)*x*(a + b*x^2))/(15*a*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (4*b*Sqrt[a*x + b*x^3])/(15*x^3) -
(8*b^2*Sqrt[a*x + b*x^3])/(15*a*x) - (2*(a*x + b*x^3)^(3/2))/(9*x^6) - (8*b^(9/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x
)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(15*a^(3/4)*S
qrt[a*x + b*x^3]) + (4*b^(9/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*Ellipti
cF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(15*a^(3/4)*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 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}+\frac {1}{3} (2 b) \int \frac {\sqrt {a x+b x^3}}{x^4} \, dx \\ & = -\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}+\frac {1}{15} \left (4 b^2\right ) \int \frac {1}{x \sqrt {a x+b x^3}} \, dx \\ & = -\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}+\frac {\left (4 b^3\right ) \int \frac {x}{\sqrt {a x+b x^3}} \, dx}{15 a} \\ & = -\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}+\frac {\left (4 b^3 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2}} \, dx}{15 a \sqrt {a x+b x^3}} \\ & = -\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}+\frac {\left (8 b^3 \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{15 a \sqrt {a x+b x^3}} \\ & = -\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}+\frac {\left (8 b^{5/2} \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{15 \sqrt {a} \sqrt {a x+b x^3}}-\frac {\left (8 b^{5/2} \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{15 \sqrt {a} \sqrt {a x+b x^3}} \\ & = \frac {8 b^{5/2} x \left (a+b x^2\right )}{15 a \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {4 b \sqrt {a x+b x^3}}{15 x^3}-\frac {8 b^2 \sqrt {a x+b x^3}}{15 a x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{9 x^6}-\frac {8 b^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a x+b x^3}}+\frac {4 b^{9/4} \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 )}{15 a^{3/4} \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 = 0.18 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {2 a \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{2},-\frac {5}{4},-\frac {b x^2}{a}\right )}{9 x^5 \sqrt {1+\frac {b x^2}{a}}} \]

[In]

Integrate[(a*x + b*x^3)^(3/2)/x^7,x]

[Out]

(-2*a*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((b*x^2)/a)])/(9*x^5*Sqrt[1 + (b*x^2)/a])

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.69

method	result	size
risch	\(-\frac {2 \left (b \,x^{2}+a \right ) \left (12 b^{2} x^{4}+11 a b \,x^{2}+5 a^{2}\right )}{45 x^{4} \sqrt {x \left (b \,x^{2}+a \right )}\, a}+\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 a \sqrt {b \,x^{3}+a x}}\)	\(210\)
default	\(-\frac {2 a \sqrt {b \,x^{3}+a x}}{9 x^{5}}-\frac {22 b \sqrt {b \,x^{3}+a x}}{45 x^{3}}-\frac {8 \left (b \,x^{2}+a \right ) b^{2}}{15 a \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 a \sqrt {b \,x^{3}+a x}}\)	\(223\)
elliptic	\(-\frac {2 a \sqrt {b \,x^{3}+a x}}{9 x^{5}}-\frac {22 b \sqrt {b \,x^{3}+a x}}{45 x^{3}}-\frac {8 \left (b \,x^{2}+a \right ) b^{2}}{15 a \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 a \sqrt {b \,x^{3}+a x}}\)	\(223\)

[In]

int((b*x^3+a*x)^(3/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

-2/45*(b*x^2+a)*(12*b^2*x^4+11*a*b*x^2+5*a^2)/x^4/(x*(b*x^2+a))^(1/2)/a+4/15*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)*(-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*Elli
pticF(((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 = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {2 \, {\left (12 \, b^{\frac {5}{2}} x^{5} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left (12 \, b^{2} x^{4} + 11 \, a b x^{2} + 5 \, a^{2}\right )} \sqrt {b x^{3} + a x}\right )}}{45 \, a x^{5}} \]

[In]

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

[Out]

-2/45*(12*b^(5/2)*x^5*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)) + (12*b^2*x^4 + 11*a*b*x^2
+ 5*a^2)*sqrt(b*x^3 + a*x))/(a*x^5)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^7, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^7, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (b\,x^3+a\,x\right )}^{3/2}}{x^7} \,d x \]

[In]

int((a*x + b*x^3)^(3/2)/x^7,x)

[Out]

int((a*x + b*x^3)^(3/2)/x^7, x)