\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{9/2}} \, dx\) [370]

Optimal result

Integrand size = 21, antiderivative size = 286 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\frac {24 b \sqrt {c} x^{3/2} \left (b+c x^2\right )}{5 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}}-\frac {24 b^{5/4} \sqrt [4]{c} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {12 b^{5/4} \sqrt [4]{c} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}} \]

[Out]

-2*(c*x^4+b*x^2)^(3/2)/x^(7/2)+24/5*b*x^(3/2)*(c*x^2+b)*c^(1/2)/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)+12/5*c
*x^(1/2)*(c*x^4+b*x^2)^(1/2)-24/5*b^(5/4)*c^(1/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arc
tan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2)
)*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/(c*x^4+b*x^2)^(1/2)+12/5*b^(5/4)*c^(1/4)*x*(cos(2*arctan(c^(1/4)*x^(
1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))
),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2045, 2046, 2057, 335, 311, 226, 1210} \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\frac {12 b^{5/4} \sqrt [4]{c} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}}-\frac {24 b^{5/4} \sqrt [4]{c} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}+\frac {24 b \sqrt {c} x^{3/2} \left (b+c x^2\right )}{5 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}} \]

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(9/2),x]

[Out]

(24*b*Sqrt[c]*x^(3/2)*(b + c*x^2))/(5*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) + (12*c*Sqrt[x]*Sqrt[b*x^2 +
c*x^4])/5 - (2*(b*x^2 + c*x^4)^(3/2))/x^(7/2) - (24*b^(5/4)*c^(1/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(
Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(5*Sqrt[b*x^2 + c*x^4]) + (12*b^(
5/4)*c^(1/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqr
t[x])/b^(1/4)], 1/2])/(5*Sqrt[b*x^2 + c*x^4])

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 2046

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 + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*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 (b x^2+c x^4\right )^{3/2}}{x^{7/2}}+(6 c) \int \frac {\sqrt {b x^2+c x^4}}{\sqrt {x}} \, dx \\ & = \frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}}+\frac {1}{5} (12 b c) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx \\ & = \frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}}+\frac {\left (12 b c x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{5 \sqrt {b x^2+c x^4}} \\ & = \frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}}+\frac {\left (24 b c x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {b x^2+c x^4}} \\ & = \frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}}+\frac {\left (24 b^{3/2} \sqrt {c} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {b x^2+c x^4}}-\frac {\left (24 b^{3/2} \sqrt {c} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {b x^2+c x^4}} \\ & = \frac {24 b \sqrt {c} x^{3/2} \left (b+c x^2\right )}{5 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {12}{5} c \sqrt {x} \sqrt {b x^2+c x^4}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{x^{7/2}}-\frac {24 b^{5/4} \sqrt [4]{c} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {12 b^{5/4} \sqrt [4]{c} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.20 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},-\frac {c x^2}{b}\right )}{x^{3/2} \sqrt {1+\frac {c x^2}{b}}} \]

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(9/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (12 b^{2} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 b^{2} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+c^{2} x^{4}-4 b c \,x^{2}-5 b^{2}\right )}{5 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2}}\)	\(216\)
risch	\(-\frac {2 \left (-c \,x^{2}+5 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{5 x^{\frac {3}{2}}}+\frac {12 b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{5 \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(222\)

[In]

int((c*x^4+b*x^2)^(3/2)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/5*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(12*b^2*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b
*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2
*2^(1/2))-6*b^2*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c
/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+c^2*x^4-4*b*c*x^2-5*b^2)

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.21 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=-\frac {2 \, {\left (12 \, b \sqrt {c} x^{2} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 5 \, b\right )} \sqrt {x}\right )}}{5 \, x^{2}} \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

-2/5*(12*b*sqrt(c)*x^2*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) - sqrt(c*x^4 + b*x^2)*(c*
x^2 - 5*b)*sqrt(x))/x^2

Sympy [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {9}{2}}}\, dx \]

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**(9/2),x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**(9/2), x)

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {9}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(9/2), x)

Giac [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {9}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(9/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{9/2}} \,d x \]

[In]

int((b*x^2 + c*x^4)^(3/2)/x^(9/2),x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^(9/2), x)