3.4.76 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^{21/2}} \, dx\) [376]

Optimal. Leaf size=350 \[ -\frac {8 c^{7/2} x^{3/2} \left (b+c x^2\right )}{65 b^2 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}+\frac {8 c^{13/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{7/4} \sqrt {b x^2+c x^4}}-\frac {4 c^{13/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{7/4} \sqrt {b x^2+c x^4}} \]

[Out]

-2/13*(c*x^4+b*x^2)^(3/2)/x^(19/2)-8/65*c^(7/2)*x^(3/2)*(c*x^2+b)/b^2/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)-
4/39*c*(c*x^4+b*x^2)^(1/2)/x^(11/2)-8/195*c^2*(c*x^4+b*x^2)^(1/2)/b/x^(7/2)+8/65*c^3*(c*x^4+b*x^2)^(1/2)/b^2/x
^(3/2)+8/65*c^(13/4)*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)))
*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)/b^(7/4)/(c*x^4+b*x^2)^(1/2)-4/65*c^(13/4)*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^(7/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2045, 2050, 2057, 335, 311, 226, 1210} \begin {gather*} -\frac {4 c^{13/4} 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 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{7/4} \sqrt {b x^2+c x^4}}+\frac {8 c^{13/4} 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 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{7/4} \sqrt {b x^2+c x^4}}-\frac {8 c^{7/2} x^{3/2} \left (b+c x^2\right )}{65 b^2 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(21/2),x]

[Out]

(-8*c^(7/2)*x^(3/2)*(b + c*x^2))/(65*b^2*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (4*c*Sqrt[b*x^2 + c*x^4]
)/(39*x^(11/2)) - (8*c^2*Sqrt[b*x^2 + c*x^4])/(195*b*x^(7/2)) + (8*c^3*Sqrt[b*x^2 + c*x^4])/(65*b^2*x^(3/2)) -
(2*(b*x^2 + c*x^4)^(3/2))/(13*x^(19/2)) + (8*c^(13/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqr
t[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(65*b^(7/4)*Sqrt[b*x^2 + c*x^4]) - (4*c^(13/4)
*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])/(65*b^(7/4)*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 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*} \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{21/2}} \, dx &=-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}+\frac {1}{13} (6 c) \int \frac {\sqrt {b x^2+c x^4}}{x^{13/2}} \, dx\\ &=-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}+\frac {1}{39} \left (4 c^2\right ) \int \frac {1}{x^{5/2} \sqrt {b x^2+c x^4}} \, dx\\ &=-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}-\frac {\left (4 c^3\right ) \int \frac {1}{\sqrt {x} \sqrt {b x^2+c x^4}} \, dx}{65 b}\\ &=-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}-\frac {\left (4 c^4\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{65 b^2}\\ &=-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}-\frac {\left (4 c^4 x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{65 b^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}-\frac {\left (8 c^4 x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{65 b^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}-\frac {\left (8 c^{7/2} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{65 b^{3/2} \sqrt {b x^2+c x^4}}+\frac {\left (8 c^{7/2} 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 )}{65 b^{3/2} \sqrt {b x^2+c x^4}}\\ &=-\frac {8 c^{7/2} x^{3/2} \left (b+c x^2\right )}{65 b^2 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 c \sqrt {b x^2+c x^4}}{39 x^{11/2}}-\frac {8 c^2 \sqrt {b x^2+c x^4}}{195 b x^{7/2}}+\frac {8 c^3 \sqrt {b x^2+c x^4}}{65 b^2 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{13 x^{19/2}}+\frac {8 c^{13/4} 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 )}{65 b^{7/4} \sqrt {b x^2+c x^4}}-\frac {4 c^{13/4} 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 )}{65 b^{7/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 58, normalized size = 0.17 \begin {gather*} -\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac {13}{4},-\frac {3}{2};-\frac {9}{4};-\frac {c x^2}{b}\right )}{13 x^{15/2} \sqrt {1+\frac {c x^2}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(21/2),x]

[Out]

(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-13/4, -3/2, -9/4, -((c*x^2)/b)])/(13*x^(15/2)*Sqrt[1 + (c*x^2)/
b])

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Maple [A]
time = 0.12, size = 250, normalized size = 0.71

method	result	size
default	\(-\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (12 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{3} x^{6}-6 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{3} x^{6}-12 c^{4} x^{8}-8 b \,c^{3} x^{6}+29 b^{2} c^{2} x^{4}+40 b^{3} c \,x^{2}+15 b^{4}\right )}{195 x^{\frac {19}{2}} \left (c \,x^{2}+b \right )^{2} b^{2}}\)	\(250\)
risch	\(-\frac {2 \left (-12 c^{3} x^{6}+4 b \,c^{2} x^{4}+25 b^{2} c \,x^{2}+15 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{195 x^{\frac {15}{2}} b^{2}}-\frac {4 c^{3} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{65 b^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(252\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^(21/2),x,method=_RETURNVERBOSE)

[Out]

-2/195*(c*x^4+b*x^2)^(3/2)/x^(19/2)/(c*x^2+b)^2*(12*((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))*b*c^3*x^6-6*((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))*b*c^3*x^6-12*c^4*x^8
-8*b*c^3*x^6+29*b^2*c^2*x^4+40*b^3*c*x^2+15*b^4)/b^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(21/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(21/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 83, normalized size = 0.24 \begin {gather*} \frac {2 \, {\left (12 \, c^{\frac {7}{2}} x^{8} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (12 \, c^{3} x^{6} - 4 \, b c^{2} x^{4} - 25 \, b^{2} c x^{2} - 15 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{195 \, b^{2} x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(21/2),x, algorithm="fricas")

[Out]

2/195*(12*c^(7/2)*x^8*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) + (12*c^3*x^6 - 4*b*c^2*x^
4 - 25*b^2*c*x^2 - 15*b^3)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^2*x^8)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**(21/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(21/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(21/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{21/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^(21/2),x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^(21/2), x)

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