3.4.91 \(\int \frac {x^{17/2}}{(b x^2+c x^4)^{3/2}} \, dx\) [391]

Optimal. Leaf size=174 \[ -\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{14 c^{13/4} \sqrt {b x^2+c x^4}} \]

[Out]

-x^(11/2)/c/(c*x^4+b*x^2)^(1/2)+9/7*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2-15/7*b*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)+15/
14*b^(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)/c^(13/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2047, 2049, 2057, 335, 226} \begin {gather*} \frac {15 b^{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 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{14 c^{13/4} \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x^(11/2)/(c*Sqrt[b*x^2 + c*x^4])) - (15*b*Sqrt[b*x^2 + c*x^4])/(7*c^3*Sqrt[x]) + (9*x^(3/2)*Sqrt[b*x^2 + c*x
^4])/(7*c^2) + (15*b^(7/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTa
n[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(14*c^(13/4)*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 2047

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1))), x] - Dist[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1))), I
nt[(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] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 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*} \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}+\frac {9 \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx}{2 c}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}-\frac {(45 b) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{14 c^2}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {\left (15 b^2\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{14 c^3}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {\left (15 b^2 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{14 c^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {\left (15 b^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 c^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {15 b^{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 )}{14 c^{13/4} \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 = 10.04, size = 86, normalized size = 0.49 \begin {gather*} \frac {x^{3/2} \left (-15 b^2-6 b c x^2+2 c^2 x^4+15 b^2 \sqrt {1+\frac {c x^2}{b}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{b}\right )\right )}{7 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^(3/2)*(-15*b^2 - 6*b*c*x^2 + 2*c^2*x^4 + 15*b^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x
^2)/b)]))/(7*c^3*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.14, size = 144, normalized size = 0.83

method	result	size
default	\(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (15 b^{2} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+4 c^{3} x^{5}-12 b \,c^{2} x^{3}-30 b^{2} c x \right )}{14 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{4}}\)	\(144\)
risch	\(-\frac {2 \left (-c \,x^{2}+4 b \right ) \left (c \,x^{2}+b \right ) x^{\frac {3}{2}}}{7 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b^{2} \left (\frac {11 \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}-7 b \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{7 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(318\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/14/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(15*b^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))+4*c^3*x^5-12*b*c^2*x^3-30*b^2*c*x)/c^4

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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(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 88, normalized size = 0.51 \begin {gather*} \frac {15 \, {\left (b^{2} c x^{3} + b^{3} x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (2 \, c^{3} x^{4} - 6 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{7 \, {\left (c^{5} x^{3} + b c^{4} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

1/7*(15*(b^2*c*x^3 + b^3*x)*sqrt(c)*weierstrassPInverse(-4*b/c, 0, x) + (2*c^3*x^4 - 6*b*c^2*x^2 - 15*b^2*c)*s
qrt(c*x^4 + b*x^2)*sqrt(x))/(c^5*x^3 + b*c^4*x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(17/2)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7771 deep

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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(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{17/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^(17/2)/(b*x^2 + c*x^4)^(3/2), x)

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