3.4.67 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^{3/2}} \, dx\) [367]

Optimal. Leaf size=173 \[ \frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}-\frac {4 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 c^{5/4} \sqrt {b x^2+c x^4}} \]

[Out]

2/11*(c*x^4+b*x^2)^(3/2)/x^(1/2)+12/77*b*x^(3/2)*(c*x^4+b*x^2)^(1/2)+8/77*b^2*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)-4/
77*b^(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)))*Elliptic
F(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^(5/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 173, 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 = {2046, 2049, 2057, 335, 226} \begin {gather*} -\frac {4 b^{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 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 c^{5/4} \sqrt {b x^2+c x^4}}+\frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(3/2),x]

[Out]

(8*b^2*Sqrt[b*x^2 + c*x^4])/(77*c*Sqrt[x]) + (12*b*x^(3/2)*Sqrt[b*x^2 + c*x^4])/77 + (2*(b*x^2 + c*x^4)^(3/2))
/(11*Sqrt[x]) - (4*b^(11/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcT
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(77*c^(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 2046

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 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 {\left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx &=\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}+\frac {1}{11} (6 b) \int \sqrt {x} \sqrt {b x^2+c x^4} \, dx\\ &=\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}+\frac {1}{77} \left (12 b^2\right ) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}-\frac {\left (4 b^3\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{77 c}\\ &=\frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}-\frac {\left (4 b^3 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{77 c \sqrt {b x^2+c x^4}}\\ &=\frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}-\frac {\left (8 b^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 c \sqrt {b x^2+c x^4}}\\ &=\frac {8 b^2 \sqrt {b x^2+c x^4}}{77 c \sqrt {x}}+\frac {12}{77} b x^{3/2} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}-\frac {4 b^{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 c^{5/4} \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 = 10.03, size = 90, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}}-b^2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{b}\right )\right )}{11 c \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 157, normalized size = 0.91

method	result	size
default	\(-\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-7 c^{4} x^{7}+2 b^{3} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-20 b \,c^{3} x^{5}-17 b^{2} c^{2} x^{3}-4 b^{3} c x \right )}{77 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{2}}\)	\(157\)
risch	\(\frac {2 \left (7 c^{2} x^{4}+13 b c \,x^{2}+4 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{77 \sqrt {x}\, c}-\frac {4 b^{3} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{77 c^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(-7*c^4*x^7+2*b^3*(-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))-20*b*c^3*x^5-17*b^2*c^2*x^3-4*b^3*c*x)/c^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^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 69, normalized size = 0.40 \begin {gather*} -\frac {2 \, {\left (4 \, b^{3} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (7 \, c^{3} x^{4} + 13 \, b c^{2} x^{2} + 4 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{77 \, c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(4*b^3*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) - (7*c^3*x^4 + 13*b*c^2*x^2 + 4*b^2*c)*sqrt(c*x^4 + b
*x^2)*sqrt(x))/(c^2*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^(3/2),x, algorithm="giac")

[Out]

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

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

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^(3/2), x)

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