3.369 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac {4 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{c} \sqrt {b x^2+c x^4}}+\frac {4 b \sqrt {b x^2+c x^4}}{7 \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2021, 2032, 329, 220} \[ \frac {4 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{c} \sqrt {b x^2+c x^4}}+\frac {4 b \sqrt {b x^2+c x^4}}{7 \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(7/2),x]

[Out]

(4*b*Sqrt[b*x^2 + c*x^4])/(7*Sqrt[x]) + (2*(b*x^2 + c*x^4)^(3/2))/(7*x^(5/2)) + (4*b^(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*c^(1/4)
*Sqrt[b*x^2 + c*x^4])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

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 2021

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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^{7/2}} \, dx &=\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}+\frac {1}{7} (6 b) \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx\\ &=\frac {4 b \sqrt {b x^2+c x^4}}{7 \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}+\frac {1}{7} \left (4 b^2\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {4 b \sqrt {b x^2+c x^4}}{7 \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}+\frac {\left (4 b^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 b \sqrt {b x^2+c x^4}}{7 \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}+\frac {\left (8 b^2 x \sqrt {b+c x^2}\right ) \operatorname {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 b \sqrt {b x^2+c x^4}}{7 \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{5/2}}+\frac {4 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 )}{7 \sqrt [4]{c} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.39 \[ \frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{b}\right )}{\sqrt {x} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(7/2),x]

[Out]

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

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}} {\left (c x^{2} + b\right )}}{x^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*(c*x^2 + b)/x^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(7/2), x)

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maple [A]  time = 0.02, size = 145, normalized size = 1.01 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (c^{3} x^{5}+4 b \,c^{2} x^{3}+3 b^{2} c x +2 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, b^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{7 \left (c \,x^{2}+b \right )^{2} c \,x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^(7/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^(7/2),x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**(7/2),x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**(7/2), x)

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