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

Optimal. Leaf size=99 \[ \frac {2 \left (a^2 x^2+b^2\right )^{3/4} \left (2 a^2 x^2-7 a b+2 b^2\right )}{7 a}-3 b^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )+3 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {446, 80, 50, 63, 298, 203, 206} \begin {gather*} -2 b \left (a^2 x^2+b^2\right )^{3/4}+\frac {4 \left (a^2 x^2+b^2\right )^{7/4}}{7 a}-3 b^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )+3 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 203

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

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 298

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

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]]

Rubi steps

\begin {align*} \int \frac {\left (-3 b+2 a x^2\right ) \left (b^2+a^2 x^2\right )^{3/4}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(-3 b+2 a x) \left (b^2+a^2 x\right )^{3/4}}{x} \, dx,x,x^2\right )\\ &=\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-\frac {1}{2} (3 b) \operatorname {Subst}\left (\int \frac {\left (b^2+a^2 x\right )^{3/4}}{x} \, dx,x,x^2\right )\\ &=-2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-\frac {1}{2} \left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{b^2+a^2 x}} \, dx,x,x^2\right )\\ &=-2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{-\frac {b^2}{a^2}+\frac {x^4}{a^2}} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )}{a^2}\\ &=-2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}+\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )-\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )\\ &=-2 b \left (b^2+a^2 x^2\right )^{3/4}+\frac {4 \left (b^2+a^2 x^2\right )^{7/4}}{7 a}-3 b^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )+3 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 99, normalized size = 1.00 \begin {gather*} \frac {2 \left (a^2 x^2+b^2\right )^{3/4} \left (2 a^2 x^2-7 a b+2 b^2\right )}{7 a}-3 b^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )+3 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 99, normalized size = 1.00 \begin {gather*} \frac {2 \left (b^2+a^2 x^2\right )^{3/4} \left (-7 a b+2 b^2+2 a^2 x^2\right )}{7 a}-3 b^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )+3 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 299, normalized size = 3.02 \begin {gather*} \left [-\frac {42 \, a b^{\frac {5}{2}} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) - 21 \, a b^{\frac {5}{2}} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b^{\frac {3}{2}} + 2 \, \sqrt {a^{2} x^{2} + b^{2}} b + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {b}}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} x^{2} - 7 \, a b + 2 \, b^{2}\right )} {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}}}{14 \, a}, -\frac {42 \, a \sqrt {-b} b^{2} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b}}{b}\right ) - 21 \, a \sqrt {-b} b^{2} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b} b - 2 \, \sqrt {a^{2} x^{2} + b^{2}} b - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {-b}}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} x^{2} - 7 \, a b + 2 \, b^{2}\right )} {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}}}{14 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/14*(42*a*b^(5/2)*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b)) - 21*a*b^(5/2)*log((a^2*x^2 + 2*b^2 + 2*(a^2*x^2 +
b^2)^(1/4)*b^(3/2) + 2*sqrt(a^2*x^2 + b^2)*b + 2*(a^2*x^2 + b^2)^(3/4)*sqrt(b))/x^2) - 4*(2*a^2*x^2 - 7*a*b +
2*b^2)*(a^2*x^2 + b^2)^(3/4))/a, -1/14*(42*a*sqrt(-b)*b^2*arctan((a^2*x^2 + b^2)^(1/4)*sqrt(-b)/b) - 21*a*sqrt
(-b)*b^2*log((a^2*x^2 + 2*b^2 + 2*(a^2*x^2 + b^2)^(1/4)*sqrt(-b)*b - 2*sqrt(a^2*x^2 + b^2)*b - 2*(a^2*x^2 + b^
2)^(3/4)*sqrt(-b))/x^2) - 4*(2*a^2*x^2 - 7*a*b + 2*b^2)*(a^2*x^2 + b^2)^(3/4))/a]

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giac [A]  time = 0.30, size = 97, normalized size = 0.98 \begin {gather*} -\frac {3 \, b^{3} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {-b}}\right )}{\sqrt {-b}} - 3 \, b^{\frac {5}{2}} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) - \frac {2 \, {\left (7 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} a^{7} b - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {7}{4}} a^{6}\right )}}{7 \, a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (2 a \,x^{2}-3 b \right ) \left (a^{2} x^{2}+b^{2}\right )^{\frac {3}{4}}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 107, normalized size = 1.08 \begin {gather*} -\frac {1}{2} \, {\left (6 \, b^{\frac {3}{2}} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) + 3 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b} + {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}\right ) + 4 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}}\right )} b + \frac {4 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {7}{4}}}{7 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 81, normalized size = 0.82 \begin {gather*} 3\,b^{5/2}\,\mathrm {atanh}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )-3\,b^{5/2}\,\mathrm {atan}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )-2\,b\,{\left (a^2\,x^2+b^2\right )}^{3/4}+\frac {4\,{\left (a^2\,x^2+b^2\right )}^{7/4}}{7\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b^(5/2)*atanh((b^2 + a^2*x^2)^(1/4)/b^(1/2)) - 3*b^(5/2)*atan((b^2 + a^2*x^2)^(1/4)/b^(1/2)) - 2*b*(b^2 + a^
2*x^2)^(3/4) + (4*(b^2 + a^2*x^2)^(7/4))/(7*a)

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sympy [A]  time = 9.77, size = 88, normalized size = 0.89 \begin {gather*} \frac {3 a^{\frac {3}{2}} 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^{2} e^{i \pi }}{a^{2} x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{4}\right )} + 2 a \left (\begin {cases} \frac {x^{2} \left (b^{2}\right )^{\frac {3}{4}}}{2} & \text {for}\: a^{2} = 0 \\\frac {2 \left (a^{2} x^{2} + b^{2}\right )^{\frac {7}{4}}}{7 a^{2}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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