3.365 \(\int \sqrt {x} (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=203 \[ \frac {4 b^{15/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}}-\frac {8 b^3 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2} \]

[Out]

2/15*x^(3/2)*(c*x^4+b*x^2)^(3/2)+8/385*b^2*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c+4/55*b*x^(7/2)*(c*x^4+b*x^2)^(1/2)-8/
231*b^3*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/2)+4/231*b^(15/4)*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^(9/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2021, 2024, 2032, 329, 220} \[ -\frac {8 b^3 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {4 b^{15/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}}+\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-8*b^3*Sqrt[b*x^2 + c*x^4])/(231*c^2*Sqrt[x]) + (8*b^2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(385*c) + (4*b*x^(7/2)*Sq
rt[b*x^2 + c*x^4])/55 + (2*x^(3/2)*(b*x^2 + c*x^4)^(3/2))/15 + (4*b^(15/4)*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])/(231*c^(9/4)*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 2024

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 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 \sqrt {x} \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}+\frac {1}{5} (2 b) \int x^{5/2} \sqrt {b x^2+c x^4} \, dx\\ &=\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}+\frac {1}{55} \left (4 b^2\right ) \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}-\frac {\left (4 b^3\right ) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{77 c}\\ &=-\frac {8 b^3 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}+\frac {\left (4 b^4\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 c^2}\\ &=-\frac {8 b^3 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}+\frac {\left (4 b^4 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{231 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {8 b^3 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}+\frac {\left (8 b^4 x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{231 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {8 b^3 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {8 b^2 x^{3/2} \sqrt {b x^2+c x^4}}{385 c}+\frac {4}{55} b x^{7/2} \sqrt {b x^2+c x^4}+\frac {2}{15} x^{3/2} \left (b x^2+c x^4\right )^{3/2}+\frac {4 b^{15/4} 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 )}{231 c^{9/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 101, normalized size = 0.50 \[ \frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (5 b^3 \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{b}\right )-\left (5 b-11 c x^2\right ) \left (b+c x^2\right )^2 \sqrt {\frac {c x^2}{b}+1}\right )}{165 c^2 \sqrt {x} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*(-((5*b - 11*c*x^2)*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b]) + 5*b^3*Hypergeometric2F1[-3/2
, 1/4, 5/4, -((c*x^2)/b)]))/(165*c^2*Sqrt[x]*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} \sqrt {x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)*x^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)*x^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 168, normalized size = 0.83 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (77 c^{5} x^{9}+196 b \,c^{4} x^{7}+131 b^{2} c^{3} x^{5}-8 b^{3} c^{2} x^{3}-20 b^{4} c x +10 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, b^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{1155 \left (c \,x^{2}+b \right )^{2} c^{3} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)*x^(1/2),x)

[Out]

2/1155*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(77*x^9*c^5+196*x^7*b*c^4+10*b^4*(-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))+131*x^5*b^2*c^3-8*x^3*b^3*c^2-20*x*b^4*c)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)*x^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^(1/2)*(b*x^2 + c*x^4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)*x**(1/2),x)

[Out]

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

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