\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{19/2}} \, dx\) [375]

Optimal result

Integrand size = 21, antiderivative size = 173 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=-\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{77 b x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}-\frac {4 c^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 b^{5/4} \sqrt {b x^2+c x^4}} \]

[Out]

-2/11*(c*x^4+b*x^2)^(3/2)/x^(17/2)-12/77*c*(c*x^4+b*x^2)^(1/2)/x^(9/2)-8/77*c^2*(c*x^4+b*x^2)^(1/2)/b/x^(5/2)-
4/77*c^(11/4)*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)))*Ellipt
icF(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^(5/4)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2045, 2050, 2057, 335, 226} \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=-\frac {4 c^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{77 b x^{5/2}}-\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}} \]

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(19/2),x]

[Out]

(-12*c*Sqrt[b*x^2 + c*x^4])/(77*x^(9/2)) - (8*c^2*Sqrt[b*x^2 + c*x^4])/(77*b*x^(5/2)) - (2*(b*x^2 + c*x^4)^(3/
2))/(11*x^(17/2)) - (4*c^(11/4)*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])/(77*b^(5/4)*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 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 (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}+\frac {1}{11} (6 c) \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx \\ & = -\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}+\frac {1}{77} \left (12 c^2\right ) \int \frac {1}{x^{3/2} \sqrt {b x^2+c x^4}} \, dx \\ & = -\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{77 b x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}-\frac {\left (4 c^3\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{77 b} \\ & = -\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{77 b x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}-\frac {\left (4 c^3 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{77 b \sqrt {b x^2+c x^4}} \\ & = -\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{77 b x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}-\frac {\left (8 c^3 x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{77 b \sqrt {b x^2+c x^4}} \\ & = -\frac {12 c \sqrt {b x^2+c x^4}}{77 x^{9/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{77 b x^{5/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 x^{17/2}}-\frac {4 c^{11/4} 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 )}{77 b^{5/4} \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 = 10.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.34 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {3}{2},-\frac {7}{4},-\frac {c x^2}{b}\right )}{11 x^{13/2} \sqrt {1+\frac {c x^2}{b}}} \]

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(19/2),x]

[Out]

(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-11/4, -3/2, -7/4, -((c*x^2)/b)])/(11*x^(13/2)*Sqrt[1 + (c*x^2)/
b])

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c^{2} x^{5}+4 c^{3} x^{6}+17 b \,c^{2} x^{4}+20 b^{2} c \,x^{2}+7 b^{3}\right )}{77 x^{\frac {17}{2}} \left (c \,x^{2}+b \right )^{2} b}\)	\(156\)
risch	\(-\frac {2 \left (4 c^{2} x^{4}+13 b c \,x^{2}+7 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{77 x^{\frac {13}{2}} b}-\frac {4 c^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{77 b \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(191\)

[In]

int((c*x^4+b*x^2)^(3/2)/x^(19/2),x,method=_RETURNVERBOSE)

[Out]

-2/77*(c*x^4+b*x^2)^(3/2)/x^(17/2)/(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^2*x^5+4*c^3*x^6+17*b*c^2*x^4+20*b^2*c*x^2+7*b^3)/b

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.37 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=-\frac {2 \, {\left (4 \, c^{\frac {5}{2}} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (4 \, c^{2} x^{4} + 13 \, b c x^{2} + 7 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{77 \, b x^{7}} \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(19/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**(19/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {19}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(19/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(19/2), x)

Giac [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {19}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(19/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(19/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{19/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{19/2}} \,d x \]

[In]

int((b*x^2 + c*x^4)^(3/2)/x^(19/2),x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^(19/2), x)