3.2.73 \(\int \frac {\sqrt {b x^{2/3}+a x}}{x^3} \, dx\) [173]

Optimal. Leaf size=178 \[ -\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}+\frac {7 a^2 \sqrt {b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac {7 a^3 \sqrt {b x^{2/3}+a x}}{64 b^3 x}+\frac {21 a^4 \sqrt {b x^{2/3}+a x}}{128 b^4 x^{2/3}}-\frac {21 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{128 b^{9/2}} \]

[Out]

-21/128*a^5*arctanh(x^(1/3)*b^(1/2)/(b*x^(2/3)+a*x)^(1/2))/b^(9/2)-3/5*(b*x^(2/3)+a*x)^(1/2)/x^2-3/40*a*(b*x^(
2/3)+a*x)^(1/2)/b/x^(5/3)+7/80*a^2*(b*x^(2/3)+a*x)^(1/2)/b^2/x^(4/3)-7/64*a^3*(b*x^(2/3)+a*x)^(1/2)/b^3/x+21/1
28*a^4*(b*x^(2/3)+a*x)^(1/2)/b^4/x^(2/3)

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Rubi [A]
time = 0.20, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2045, 2050, 2054, 212} \begin {gather*} -\frac {21 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{128 b^{9/2}}+\frac {21 a^4 \sqrt {a x+b x^{2/3}}}{128 b^4 x^{2/3}}-\frac {7 a^3 \sqrt {a x+b x^{2/3}}}{64 b^3 x}+\frac {7 a^2 \sqrt {a x+b x^{2/3}}}{80 b^2 x^{4/3}}-\frac {3 a \sqrt {a x+b x^{2/3}}}{40 b x^{5/3}}-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^(2/3) + a*x]/x^3,x]

[Out]

(-3*Sqrt[b*x^(2/3) + a*x])/(5*x^2) - (3*a*Sqrt[b*x^(2/3) + a*x])/(40*b*x^(5/3)) + (7*a^2*Sqrt[b*x^(2/3) + a*x]
)/(80*b^2*x^(4/3)) - (7*a^3*Sqrt[b*x^(2/3) + a*x])/(64*b^3*x) + (21*a^4*Sqrt[b*x^(2/3) + a*x])/(128*b^4*x^(2/3
)) - (21*a^5*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(128*b^(9/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {\sqrt {b x^{2/3}+a x}}{x^3} \, dx &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}+\frac {1}{10} a \int \frac {1}{x^2 \sqrt {b x^{2/3}+a x}} \, dx\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}-\frac {\left (7 a^2\right ) \int \frac {1}{x^{5/3} \sqrt {b x^{2/3}+a x}} \, dx}{80 b}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}+\frac {7 a^2 \sqrt {b x^{2/3}+a x}}{80 b^2 x^{4/3}}+\frac {\left (7 a^3\right ) \int \frac {1}{x^{4/3} \sqrt {b x^{2/3}+a x}} \, dx}{96 b^2}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}+\frac {7 a^2 \sqrt {b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac {7 a^3 \sqrt {b x^{2/3}+a x}}{64 b^3 x}-\frac {\left (7 a^4\right ) \int \frac {1}{x \sqrt {b x^{2/3}+a x}} \, dx}{128 b^3}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}+\frac {7 a^2 \sqrt {b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac {7 a^3 \sqrt {b x^{2/3}+a x}}{64 b^3 x}+\frac {21 a^4 \sqrt {b x^{2/3}+a x}}{128 b^4 x^{2/3}}+\frac {\left (7 a^5\right ) \int \frac {1}{x^{2/3} \sqrt {b x^{2/3}+a x}} \, dx}{256 b^4}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}+\frac {7 a^2 \sqrt {b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac {7 a^3 \sqrt {b x^{2/3}+a x}}{64 b^3 x}+\frac {21 a^4 \sqrt {b x^{2/3}+a x}}{128 b^4 x^{2/3}}-\frac {\left (21 a^5\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{128 b^4}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{5 x^2}-\frac {3 a \sqrt {b x^{2/3}+a x}}{40 b x^{5/3}}+\frac {7 a^2 \sqrt {b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac {7 a^3 \sqrt {b x^{2/3}+a x}}{64 b^3 x}+\frac {21 a^4 \sqrt {b x^{2/3}+a x}}{128 b^4 x^{2/3}}-\frac {21 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{128 b^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 112, normalized size = 0.63 \begin {gather*} \frac {\sqrt {b x^{2/3}+a x} \left (-384 b^4-48 a b^3 \sqrt [3]{x}+56 a^2 b^2 x^{2/3}-70 a^3 b x+105 a^4 x^{4/3}\right )}{640 b^4 x^2}-\frac {21 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{128 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^(2/3) + a*x]*(-384*b^4 - 48*a*b^3*x^(1/3) + 56*a^2*b^2*x^(2/3) - 70*a^3*b*x + 105*a^4*x^(4/3)))/(640
*b^4*x^2) - (21*a^5*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(128*b^(9/2))

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Maple [A]
time = 0.35, size = 125, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {\sqrt {b \,x^{\frac {2}{3}}+a x}\, \left (105 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {9}{2}} b^{\frac {9}{2}}-490 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {7}{2}} b^{\frac {11}{2}}+896 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {5}{2}} b^{\frac {13}{2}}-790 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} b^{\frac {15}{2}}-105 \arctanh \left (\frac {\sqrt {b +a \,x^{\frac {1}{3}}}}{\sqrt {b}}\right ) b^{4} a^{5} x^{\frac {5}{3}}-105 \sqrt {b +a \,x^{\frac {1}{3}}}\, b^{\frac {17}{2}}\right )}{640 x^{2} \sqrt {b +a \,x^{\frac {1}{3}}}\, b^{\frac {17}{2}}}\)	\(125\)
default	\(-\frac {\sqrt {b \,x^{\frac {2}{3}}+a x}\, \left (-105 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {9}{2}} b^{\frac {9}{2}}+490 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {7}{2}} b^{\frac {11}{2}}-896 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {5}{2}} b^{\frac {13}{2}}+105 \arctanh \left (\frac {\sqrt {b +a \,x^{\frac {1}{3}}}}{\sqrt {b}}\right ) b^{4} a^{5} x^{\frac {5}{3}}+790 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} b^{\frac {15}{2}}+105 \sqrt {b +a \,x^{\frac {1}{3}}}\, b^{\frac {17}{2}}\right )}{640 x^{2} \sqrt {b +a \,x^{\frac {1}{3}}}\, b^{\frac {17}{2}}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^(2/3)+a*x)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/640*(b*x^(2/3)+a*x)^(1/2)*(-105*(b+a*x^(1/3))^(9/2)*b^(9/2)+490*(b+a*x^(1/3))^(7/2)*b^(11/2)-896*(b+a*x^(1/
3))^(5/2)*b^(13/2)+105*arctanh((b+a*x^(1/3))^(1/2)/b^(1/2))*b^4*a^5*x^(5/3)+790*(b+a*x^(1/3))^(3/2)*b^(15/2)+1
05*(b+a*x^(1/3))^(1/2)*b^(17/2))/x^2/(b+a*x^(1/3))^(1/2)/b^(17/2)

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

[Out]

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

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Fricas [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((b*x^(2/3)+a*x)^(1/2)/x^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x + b x^{\frac {2}{3}}}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.53, size = 126, normalized size = 0.71 \begin {gather*} \frac {\frac {105 \, a^{6} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{4}} + \frac {105 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} a^{6} - 490 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} a^{6} b + 896 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{6} b^{2} - 790 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{6} b^{3} - 105 \, \sqrt {a x^{\frac {1}{3}} + b} a^{6} b^{4}}{a^{5} b^{4} x^{\frac {5}{3}}}}{640 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*(105*a^6*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) + (105*(a*x^(1/3) + b)^(9/2)*a^6 - 490*(a*x
^(1/3) + b)^(7/2)*a^6*b + 896*(a*x^(1/3) + b)^(5/2)*a^6*b^2 - 790*(a*x^(1/3) + b)^(3/2)*a^6*b^3 - 105*sqrt(a*x
^(1/3) + b)*a^6*b^4)/(a^5*b^4*x^(5/3)))/a

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a\,x+b\,x^{2/3}}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(1/2)/x^3,x)

[Out]

int((a*x + b*x^(2/3))^(1/2)/x^3, x)

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