\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{15/2}} \, dx\) [373]

Optimal result

Integrand size = 21, antiderivative size = 143 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=-\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac {4 c^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {b x^2+c x^4}} \]

[Out]

-2/7*(c*x^4+b*x^2)^(3/2)/x^(13/2)-4/7*c*(c*x^4+b*x^2)^(1/2)/x^(5/2)+4/7*c^(7/4)*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)/b^(1/4)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2045, 2057, 335, 226} \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\frac {4 c^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {b x^2+c x^4}}-\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}} \]

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(15/2),x]

[Out]

(-4*c*Sqrt[b*x^2 + c*x^4])/(7*x^(5/2)) - (2*(b*x^2 + c*x^4)^(3/2))/(7*x^(13/2)) + (4*c^(7/4)*x*(Sqrt[b] + Sqrt
[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(7*b^(1/
4)*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 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 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 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}}{7 x^{13/2}}+\frac {1}{7} (6 c) \int \frac {\sqrt {b x^2+c x^4}}{x^{7/2}} \, dx \\ & = -\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac {1}{7} \left (4 c^2\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx \\ & = -\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac {\left (4 c^2 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{7 \sqrt {b x^2+c x^4}} \\ & = -\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac {\left (8 c^2 x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{7 \sqrt {b x^2+c x^4}} \\ & = -\frac {4 c \sqrt {b x^2+c x^4}}{7 x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac {4 c^{7/4} 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 )}{7 \sqrt [4]{b} \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 = 58, normalized size of antiderivative = 0.41 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {3}{2},-\frac {3}{4},-\frac {c x^2}{b}\right )}{7 x^{9/2} \sqrt {1+\frac {c x^2}{b}}} \]

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(15/2),x]

[Out]

(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-7/4, -3/2, -3/4, -((c*x^2)/b)])/(7*x^(9/2)*Sqrt[1 + (c*x^2)/b])

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}-3 c^{2} x^{4}-4 b c \,x^{2}-b^{2}\right )}{7 x^{\frac {13}{2}} \left (c \,x^{2}+b \right )^{2}}\)	\(140\)
risch	\(-\frac {2 \left (3 c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{7 x^{\frac {9}{2}}}+\frac {4 c \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{7 \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(170\)

[In]

int((c*x^4+b*x^2)^(3/2)/x^(15/2),x,method=_RETURNVERBOSE)

[Out]

2/7*(c*x^4+b*x^2)^(3/2)/x^(13/2)/(c*x^2+b)^2*(2*(-b*c)^(1/2)*((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*x^3-3*c^2*x^4-4*b*c*x^2-b^2)

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.34 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\frac {2 \, {\left (4 \, c^{\frac {3}{2}} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} {\left (3 \, c x^{2} + b\right )} \sqrt {x}\right )}}{7 \, x^{5}} \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="fricas")

[Out]

2/7*(4*c^(3/2)*x^5*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x^4 + b*x^2)*(3*c*x^2 + b)*sqrt(x))/x^5

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**(15/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(15/2), x)

Giac [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(15/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(15/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{15/2}} \,d x \]

[In]

int((b*x^2 + c*x^4)^(3/2)/x^(15/2),x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^(15/2), x)