3.182 \(\int \frac{(b x^{2/3}+a x)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=203 \[ \frac{21 a^5 \sqrt{a x+b x^{2/3}}}{512 b^4 x^{2/3}}-\frac{7 a^4 \sqrt{a x+b x^{2/3}}}{256 b^3 x}+\frac{7 a^3 \sqrt{a x+b x^{2/3}}}{320 b^2 x^{4/3}}-\frac{21 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{512 b^{9/2}}-\frac{3 a^2 \sqrt{a x+b x^{2/3}}}{160 b x^{5/3}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{20 x^2}-\frac{\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3} \]

[Out]

(-3*a*Sqrt[b*x^(2/3) + a*x])/(20*x^2) - (3*a^2*Sqrt[b*x^(2/3) + a*x])/(160*b*x^(5/3)) + (7*a^3*Sqrt[b*x^(2/3)
+ a*x])/(320*b^2*x^(4/3)) - (7*a^4*Sqrt[b*x^(2/3) + a*x])/(256*b^3*x) + (21*a^5*Sqrt[b*x^(2/3) + a*x])/(512*b^
4*x^(2/3)) - (b*x^(2/3) + a*x)^(3/2)/(2*x^3) - (21*a^6*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(512*
b^(9/2))

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Rubi [A]  time = 0.34037, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2029, 206} \[ \frac{21 a^5 \sqrt{a x+b x^{2/3}}}{512 b^4 x^{2/3}}-\frac{7 a^4 \sqrt{a x+b x^{2/3}}}{256 b^3 x}+\frac{7 a^3 \sqrt{a x+b x^{2/3}}}{320 b^2 x^{4/3}}-\frac{21 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{512 b^{9/2}}-\frac{3 a^2 \sqrt{a x+b x^{2/3}}}{160 b x^{5/3}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{20 x^2}-\frac{\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(3/2)/x^4,x]

[Out]

(-3*a*Sqrt[b*x^(2/3) + a*x])/(20*x^2) - (3*a^2*Sqrt[b*x^(2/3) + a*x])/(160*b*x^(5/3)) + (7*a^3*Sqrt[b*x^(2/3)
+ a*x])/(320*b^2*x^(4/3)) - (7*a^4*Sqrt[b*x^(2/3) + a*x])/(256*b^3*x) + (21*a^5*Sqrt[b*x^(2/3) + a*x])/(512*b^
4*x^(2/3)) - (b*x^(2/3) + a*x)^(3/2)/(2*x^3) - (21*a^6*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(512*
b^(9/2))

Rule 2020

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 2025

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 2029

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]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^4} \, dx &=-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}+\frac{1}{4} a \int \frac{\sqrt{b x^{2/3}+a x}}{x^3} \, dx\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}+\frac{1}{40} a^2 \int \frac{1}{x^2 \sqrt{b x^{2/3}+a x}} \, dx\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{160 b x^{5/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}-\frac{\left (7 a^3\right ) \int \frac{1}{x^{5/3} \sqrt{b x^{2/3}+a x}} \, dx}{320 b}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{160 b x^{5/3}}+\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{320 b^2 x^{4/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}+\frac{\left (7 a^4\right ) \int \frac{1}{x^{4/3} \sqrt{b x^{2/3}+a x}} \, dx}{384 b^2}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{160 b x^{5/3}}+\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{320 b^2 x^{4/3}}-\frac{7 a^4 \sqrt{b x^{2/3}+a x}}{256 b^3 x}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}-\frac{\left (7 a^5\right ) \int \frac{1}{x \sqrt{b x^{2/3}+a x}} \, dx}{512 b^3}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{160 b x^{5/3}}+\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{320 b^2 x^{4/3}}-\frac{7 a^4 \sqrt{b x^{2/3}+a x}}{256 b^3 x}+\frac{21 a^5 \sqrt{b x^{2/3}+a x}}{512 b^4 x^{2/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}+\frac{\left (7 a^6\right ) \int \frac{1}{x^{2/3} \sqrt{b x^{2/3}+a x}} \, dx}{1024 b^4}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{160 b x^{5/3}}+\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{320 b^2 x^{4/3}}-\frac{7 a^4 \sqrt{b x^{2/3}+a x}}{256 b^3 x}+\frac{21 a^5 \sqrt{b x^{2/3}+a x}}{512 b^4 x^{2/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}-\frac{\left (21 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt [3]{x}}{\sqrt{b x^{2/3}+a x}}\right )}{512 b^4}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{20 x^2}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{160 b x^{5/3}}+\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{320 b^2 x^{4/3}}-\frac{7 a^4 \sqrt{b x^{2/3}+a x}}{256 b^3 x}+\frac{21 a^5 \sqrt{b x^{2/3}+a x}}{512 b^4 x^{2/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}-\frac{21 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{b x^{2/3}+a x}}\right )}{512 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0463573, size = 61, normalized size = 0.3 \[ -\frac{6 a^6 \left (a \sqrt [3]{x}+b\right )^2 \sqrt{a x+b x^{2/3}} \, _2F_1\left (\frac{5}{2},7;\frac{7}{2};\frac{\sqrt [3]{x} a}{b}+1\right )}{5 b^7 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 139, normalized size = 0.7 \begin{align*}{\frac{1}{2560\,{x}^{3}} \left ( b{x}^{{\frac{2}{3}}}+ax \right ) ^{{\frac{3}{2}}} \left ( 105\,{b}^{9/2} \left ( b+a\sqrt [3]{x} \right ) ^{11/2}-595\,{b}^{11/2} \left ( b+a\sqrt [3]{x} \right ) ^{9/2}+1386\,{b}^{13/2} \left ( b+a\sqrt [3]{x} \right ) ^{7/2}-1686\,{b}^{15/2} \left ( b+a\sqrt [3]{x} \right ) ^{5/2}-595\,{b}^{17/2} \left ( b+a\sqrt [3]{x} \right ) ^{3/2}+105\,{b}^{19/2}\sqrt{b+a\sqrt [3]{x}}-105\,{\it Artanh} \left ({\frac{\sqrt{b+a\sqrt [3]{x}}}{\sqrt{b}}} \right ){b}^{4}{a}^{6}{x}^{2} \right ){b}^{-{\frac{17}{2}}} \left ( b+a\sqrt [3]{x} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2560*(b*x^(2/3)+a*x)^(3/2)*(105*b^(9/2)*(b+a*x^(1/3))^(11/2)-595*b^(11/2)*(b+a*x^(1/3))^(9/2)+1386*b^(13/2)*
(b+a*x^(1/3))^(7/2)-1686*b^(15/2)*(b+a*x^(1/3))^(5/2)-595*b^(17/2)*(b+a*x^(1/3))^(3/2)+105*b^(19/2)*(b+a*x^(1/
3))^(1/2)-105*arctanh((b+a*x^(1/3))^(1/2)/b^(1/2))*b^4*a^6*x^2)/x^3/(b+a*x^(1/3))^(3/2)/b^(17/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + b x^{\frac{2}{3}}\right )}^{\frac{3}{2}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.25095, size = 193, normalized size = 0.95 \begin{align*} \frac{\frac{105 \, a^{7} \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{11}{2}} a^{7} - 595 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{9}{2}} a^{7} b + 1386 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{7}{2}} a^{7} b^{2} - 1686 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} a^{7} b^{3} - 595 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} a^{7} b^{4} + 105 \, \sqrt{a x^{\frac{1}{3}} + b} a^{7} b^{5}}{a^{6} b^{4} x^{2}}}{2560 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2560*(105*a^7*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) + (105*(a*x^(1/3) + b)^(11/2)*a^7 - 595*(a
*x^(1/3) + b)^(9/2)*a^7*b + 1386*(a*x^(1/3) + b)^(7/2)*a^7*b^2 - 1686*(a*x^(1/3) + b)^(5/2)*a^7*b^3 - 595*(a*x
^(1/3) + b)^(3/2)*a^7*b^4 + 105*sqrt(a*x^(1/3) + b)*a^7*b^5)/(a^6*b^4*x^2))/a