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

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

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Rubi [A]  time = 0.34, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2020, 2025, 2029, 206} \begin {gather*} \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} \end {gather*}

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 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])

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]

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*}

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Mathematica [C]  time = 0.05, size = 61, normalized size = 0.30 \begin {gather*} -\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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 17.13, size = 152, normalized size = 0.75 \begin {gather*} \frac {\left (x^{2/3} \left (a \sqrt [3]{x}+b\right )\right )^{3/2} \left (\frac {\sqrt {a \sqrt [3]{x}+b} \left (105 a^5 x^{5/3}-70 a^4 b x^{4/3}+56 a^3 b^2 x-48 a^2 b^3 x^{2/3}-1664 a b^4 \sqrt [3]{x}-1280 b^5\right )}{2560 b^4 x^2}-\frac {21 a^6 \tanh ^{-1}\left (\frac {\sqrt {a \sqrt [3]{x}+b}}{\sqrt {b}}\right )}{512 b^{9/2}}\right )}{x \left (a \sqrt [3]{x}+b\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((b + a*x^(1/3))*x^(2/3))^(3/2)*((Sqrt[b + a*x^(1/3)]*(-1280*b^5 - 1664*a*b^4*x^(1/3) - 48*a^2*b^3*x^(2/3) +
56*a^3*b^2*x - 70*a^4*b*x^(4/3) + 105*a^5*x^(5/3)))/(2560*b^4*x^2) - (21*a^6*ArcTanh[Sqrt[b + a*x^(1/3)]/Sqrt[
b]])/(512*b^(9/2))))/((b + a*x^(1/3))^(3/2)*x)

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fricas [F(-1)]  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)^(3/2)/x^4,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.37, size = 143, normalized size = 0.70 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.06, size = 139, normalized size = 0.68 \begin {gather*} \frac {\left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} \left (-105 a^{6} b^{4} x^{2} \arctanh \left (\frac {\sqrt {a \,x^{\frac {1}{3}}+b}}{\sqrt {b}}\right )+105 \sqrt {a \,x^{\frac {1}{3}}+b}\, b^{\frac {19}{2}}-595 \left (a \,x^{\frac {1}{3}}+b \right )^{\frac {3}{2}} b^{\frac {17}{2}}-1686 \left (a \,x^{\frac {1}{3}}+b \right )^{\frac {5}{2}} b^{\frac {15}{2}}+1386 \left (a \,x^{\frac {1}{3}}+b \right )^{\frac {7}{2}} b^{\frac {13}{2}}-595 \left (a \,x^{\frac {1}{3}}+b \right )^{\frac {9}{2}} b^{\frac {11}{2}}+105 \left (a \,x^{\frac {1}{3}}+b \right )^{\frac {11}{2}} b^{\frac {9}{2}}\right )}{2560 \left (a \,x^{\frac {1}{3}}+b \right )^{\frac {3}{2}} b^{\frac {17}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(3/2)/x^4,x)

[Out]

int((a*x + b*x^(2/3))^(3/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x + b x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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