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

Optimal. Leaf size=151 \[ \frac {3 b^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}}{\sqrt {a x^2-b}-\sqrt {b}}\right )}{\sqrt {2}}+\frac {3 b^{7/4} \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^2-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt [4]{a x^2-b}}\right )}{\sqrt {2}}+\frac {2}{7} \left (a x^2-b\right )^{3/4} \left (2 a x^2+5 b\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 230, normalized size of antiderivative = 1.52, number of steps used = 13, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {446, 80, 50, 63, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {3 b^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}+\frac {3 b^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}+\frac {3 b^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {3 b^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2}}+2 b \left (a x^2-b\right )^{3/4}+\frac {4}{7} \left (a x^2-b\right )^{7/4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*b*(-b + a*x^2)^(3/4) + (4*(-b + a*x^2)^(7/4))/7 + (3*b^(7/4)*ArcTan[1 - (Sqrt[2]*(-b + a*x^2)^(1/4))/b^(1/4)
])/Sqrt[2] - (3*b^(7/4)*ArcTan[1 + (Sqrt[2]*(-b + a*x^2)^(1/4))/b^(1/4)])/Sqrt[2] - (3*b^(7/4)*Log[Sqrt[b] - S
qrt[2]*b^(1/4)*(-b + a*x^2)^(1/4) + Sqrt[-b + a*x^2]])/(2*Sqrt[2]) + (3*b^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*
(-b + a*x^2)^(1/4) + Sqrt[-b + a*x^2]])/(2*Sqrt[2])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 204

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

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {\left (-b+a x^2\right )^{3/4} \left (3 b+2 a x^2\right )}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(-b+a x)^{3/4} (3 b+2 a x)}{x} \, dx,x,x^2\right )\\ &=\frac {4}{7} \left (-b+a x^2\right )^{7/4}+\frac {1}{2} (3 b) \operatorname {Subst}\left (\int \frac {(-b+a x)^{3/4}}{x} \, dx,x,x^2\right )\\ &=2 b \left (-b+a x^2\right )^{3/4}+\frac {4}{7} \left (-b+a x^2\right )^{7/4}-\frac {1}{2} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{-b+a x}} \, dx,x,x^2\right )\\ &=2 b \left (-b+a x^2\right )^{3/4}+\frac {4}{7} \left (-b+a x^2\right )^{7/4}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{a}\\ &=2 b \left (-b+a x^2\right )^{3/4}+\frac {4}{7} \left (-b+a x^2\right )^{7/4}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{a}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{a}\\ &=2 b \left (-b+a x^2\right )^{3/4}+\frac {4}{7} \left (-b+a x^2\right )^{7/4}-\frac {\left (3 b^{7/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}+2 x}{-\sqrt {b}-\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{2 \sqrt {2}}-\frac {\left (3 b^{7/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}-2 x}{-\sqrt {b}+\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{2 \sqrt {2}}-\frac {1}{2} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )-\frac {1}{2} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )\\ &=2 b \left (-b+a x^2\right )^{3/4}+\frac {4}{7} \left (-b+a x^2\right )^{7/4}-\frac {3 b^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}+\frac {3 b^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}-\frac {\left (3 b^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}+\frac {\left (3 b^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}\\ &=2 b \left (-b+a x^2\right )^{3/4}+\frac {4}{7} \left (-b+a x^2\right )^{7/4}+\frac {3 b^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {3 b^{7/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {3 b^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}+\frac {3 b^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 91, normalized size = 0.60 \begin {gather*} \frac {2}{7} \left (a x^2-b\right )^{3/4} \left (2 a x^2+5 b\right )+3 b (-b)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a x^2-b}}{\sqrt [4]{-b}}\right )+3 (-b)^{7/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a x^2-b}}{\sqrt [4]{-b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-b + a*x^2)^(3/4)*(5*b + 2*a*x^2))/7 + 3*(-b)^(3/4)*b*ArcTan[(-b + a*x^2)^(1/4)/(-b)^(1/4)] + 3*(-b)^(7/4)
*ArcTanh[(-b + a*x^2)^(1/4)/(-b)^(1/4)]

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IntegrateAlgebraic [A]  time = 0.22, size = 150, normalized size = 0.99 \begin {gather*} \frac {2}{7} \left (-b+a x^2\right )^{3/4} \left (5 b+2 a x^2\right )-\frac {3 b^{7/4} \tan ^{-1}\left (\frac {-\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^2}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {2}}+\frac {3 b^{7/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{\sqrt {b}+\sqrt {-b+a x^2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-b + a*x^2)^(3/4)*(5*b + 2*a*x^2))/7 - (3*b^(7/4)*ArcTan[(-(b^(1/4)/Sqrt[2]) + Sqrt[-b + a*x^2]/(Sqrt[2]*b
^(1/4)))/(-b + a*x^2)^(1/4)])/Sqrt[2] + (3*b^(7/4)*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^2)^(1/4))/(Sqrt[b] + Sqr
t[-b + a*x^2])])/Sqrt[2]

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fricas [A]  time = 0.97, size = 173, normalized size = 1.15 \begin {gather*} \frac {2}{7} \, {\left (2 \, a x^{2} + 5 \, b\right )} {\left (a x^{2} - b\right )}^{\frac {3}{4}} + 6 \, \left (-b^{7}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left (-b^{7}\right )^{\frac {1}{4}} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} - \sqrt {\sqrt {a x^{2} - b} b^{10} - \sqrt {-b^{7}} b^{7}} \left (-b^{7}\right )^{\frac {1}{4}}}{b^{7}}\right ) - \frac {3}{2} \, \left (-b^{7}\right )^{\frac {1}{4}} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} + 27 \, \left (-b^{7}\right )^{\frac {3}{4}}\right ) + \frac {3}{2} \, \left (-b^{7}\right )^{\frac {1}{4}} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{5} - 27 \, \left (-b^{7}\right )^{\frac {3}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(2*a*x^2 + 5*b)*(a*x^2 - b)^(3/4) + 6*(-b^7)^(1/4)*arctan(-((-b^7)^(1/4)*(a*x^2 - b)^(1/4)*b^5 - sqrt(sqrt
(a*x^2 - b)*b^10 - sqrt(-b^7)*b^7)*(-b^7)^(1/4))/b^7) - 3/2*(-b^7)^(1/4)*log(27*(a*x^2 - b)^(1/4)*b^5 + 27*(-b
^7)^(3/4)) + 3/2*(-b^7)^(1/4)*log(27*(a*x^2 - b)^(1/4)*b^5 - 27*(-b^7)^(3/4))

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giac [A]  time = 0.26, size = 189, normalized size = 1.25 \begin {gather*} -\frac {3}{2} \, \sqrt {2} b^{\frac {7}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {3}{2} \, \sqrt {2} b^{\frac {7}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) + \frac {3}{4} \, \sqrt {2} b^{\frac {7}{4}} \log \left (\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right ) - \frac {3}{4} \, \sqrt {2} b^{\frac {7}{4}} \log \left (-\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right ) + \frac {4}{7} \, {\left (a x^{2} - b\right )}^{\frac {7}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {3}{4}} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*sqrt(2)*b^(7/4)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^2 - b)^(1/4))/b^(1/4)) - 3/2*sqrt(2)*b^(7/4)
*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^2 - b)^(1/4))/b^(1/4)) + 3/4*sqrt(2)*b^(7/4)*log(sqrt(2)*(a*x^2
- b)^(1/4)*b^(1/4) + sqrt(a*x^2 - b) + sqrt(b)) - 3/4*sqrt(2)*b^(7/4)*log(-sqrt(2)*(a*x^2 - b)^(1/4)*b^(1/4)
+ sqrt(a*x^2 - b) + sqrt(b)) + 4/7*(a*x^2 - b)^(7/4) + 2*(a*x^2 - b)^(3/4)*b

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{2}-b \right )^{\frac {3}{4}} \left (2 a \,x^{2}+3 b \right )}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 195, normalized size = 1.29 \begin {gather*} -\frac {1}{4} \, {\left (3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}}\right )} b - 8 \, {\left (a x^{2} - b\right )}^{\frac {3}{4}}\right )} b + \frac {4}{7} \, {\left (a x^{2} - b\right )}^{\frac {7}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^2 - b)^(1/4))/b^(1/4))/b^(1/4) + 2*sqrt(2)*arc
tan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^2 - b)^(1/4))/b^(1/4))/b^(1/4) - sqrt(2)*log(sqrt(2)*(a*x^2 - b)^(1
/4)*b^(1/4) + sqrt(a*x^2 - b) + sqrt(b))/b^(1/4) + sqrt(2)*log(-sqrt(2)*(a*x^2 - b)^(1/4)*b^(1/4) + sqrt(a*x^2
- b) + sqrt(b))/b^(1/4))*b - 8*(a*x^2 - b)^(3/4))*b + 4/7*(a*x^2 - b)^(7/4)

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mupad [B]  time = 1.24, size = 78, normalized size = 0.52 \begin {gather*} \frac {4\,{\left (a\,x^2-b\right )}^{7/4}}{7}-3\,{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )+3\,{\left (-b\right )}^{7/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )+2\,b\,{\left (a\,x^2-b\right )}^{3/4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(a*x^2 - b)^(7/4))/7 - 3*(-b)^(7/4)*atan((a*x^2 - b)^(1/4)/(-b)^(1/4)) + 3*(-b)^(7/4)*atanh((a*x^2 - b)^(1/
4)/(-b)^(1/4)) + 2*b*(a*x^2 - b)^(3/4)

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sympy [A]  time = 10.02, size = 80, normalized size = 0.53 \begin {gather*} - \frac {3 a^{\frac {3}{4}} b x^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{4}\right )} + 2 a \left (\begin {cases} \frac {x^{2} \left (- b\right )^{\frac {3}{4}}}{2} & \text {for}\: a = 0 \\\frac {2 \left (a x^{2} - b\right )^{\frac {7}{4}}}{7 a} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a**(3/4)*b*x**(3/2)*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), b*exp_polar(2*I*pi)/(a*x**2))/(2*gamma(1/4)) +
2*a*Piecewise((x**2*(-b)**(3/4)/2, Eq(a, 0)), (2*(a*x**2 - b)**(7/4)/(7*a), True))

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