3.5.29 \(\int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=71 \[ -\frac {a^2 \sqrt {c+d x}}{c x}-\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {2 b^2 \sqrt {c+d x}}{d} \]

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Rubi [A]  time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {89, 80, 63, 208} \begin {gather*} -\frac {a^2 \sqrt {c+d x}}{c x}-\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}}+\frac {2 b^2 \sqrt {c+d x}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^2*Sqrt[c + d*x]),x]

[Out]

(2*b^2*Sqrt[c + d*x])/d - (a^2*Sqrt[c + d*x])/(c*x) - (a*(4*b*c - a*d)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/c^(3/2)

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 208

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 65, normalized size = 0.92 \begin {gather*} \frac {\sqrt {c+d x} \left (2 b^2 c x-a^2 d\right )}{c d x}+\frac {a (a d-4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(a^2*d) + 2*b^2*c*x)*Sqrt[c + d*x])/(c*d*x) + (a*(-4*b*c + a*d)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/c^(3/2)

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IntegrateAlgebraic [A]  time = 0.12, size = 81, normalized size = 1.14 \begin {gather*} \frac {\sqrt {c+d x} \left (-a^2 d^2-2 b^2 c^2+2 b^2 c (c+d x)\right )}{c d^2 x}+\frac {\left (a^2 d-4 a b c\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2/(x^2*Sqrt[c + d*x]),x]

[Out]

(Sqrt[c + d*x]*(-2*b^2*c^2 - a^2*d^2 + 2*b^2*c*(c + d*x)))/(c*d^2*x) + ((-4*a*b*c + a^2*d)*ArcTanh[Sqrt[c + d*
x]/Sqrt[c]])/c^(3/2)

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fricas [A]  time = 1.31, size = 158, normalized size = 2.23 \begin {gather*} \left [-\frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt {c} x \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, b^{2} c^{2} x - a^{2} c d\right )} \sqrt {d x + c}}{2 \, c^{2} d x}, \frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (2 \, b^{2} c^{2} x - a^{2} c d\right )} \sqrt {d x + c}}{c^{2} d x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*((4*a*b*c*d - a^2*d^2)*sqrt(c)*x*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*b^2*c^2*x - a^2*c*d
)*sqrt(d*x + c))/(c^2*d*x), ((4*a*b*c*d - a^2*d^2)*sqrt(-c)*x*arctan(sqrt(d*x + c)*sqrt(-c)/c) + (2*b^2*c^2*x
- a^2*c*d)*sqrt(d*x + c))/(c^2*d*x)]

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giac [A]  time = 1.36, size = 74, normalized size = 1.04 \begin {gather*} \frac {2 \, \sqrt {d x + c} b^{2} - \frac {\sqrt {d x + c} a^{2} d}{c x} + \frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*sqrt(d*x + c)*b^2 - sqrt(d*x + c)*a^2*d/(c*x) + (4*a*b*c*d - a^2*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(
-c)*c))/d

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maple [A]  time = 0.02, size = 63, normalized size = 0.89 \begin {gather*} \frac {2 \left (\frac {\left (a d -4 b c \right ) \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}-\frac {\sqrt {d x +c}\, a}{2 c x}\right ) a d +2 \sqrt {d x +c}\, b^{2}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^2/(d*x+c)^(1/2),x)

[Out]

2/d*(b^2*(d*x+c)^(1/2)+a*d*(-1/2*a/c*(d*x+c)^(1/2)/x+1/2*(a*d-4*b*c)/c^(3/2)*arctanh((d*x+c)^(1/2)/c^(1/2))))

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maxima [A]  time = 1.98, size = 92, normalized size = 1.30 \begin {gather*} -\frac {1}{2} \, {\left (\frac {2 \, \sqrt {d x + c} a^{2}}{{\left (d x + c\right )} c - c^{2}} - \frac {4 \, \sqrt {d x + c} b^{2}}{d^{2}} - \frac {{\left (4 \, b c - a d\right )} a \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}} d}\right )} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.16, size = 59, normalized size = 0.83 \begin {gather*} \frac {2\,b^2\,\sqrt {c+d\,x}}{d}+\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (a\,d-4\,b\,c\right )}{c^{3/2}}-\frac {a^2\,\sqrt {c+d\,x}}{c\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(x^2*(c + d*x)^(1/2)),x)

[Out]

(2*b^2*(c + d*x)^(1/2))/d + (a*atanh((c + d*x)^(1/2)/c^(1/2))*(a*d - 4*b*c))/c^(3/2) - (a^2*(c + d*x)^(1/2))/(
c*x)

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sympy [A]  time = 92.93, size = 109, normalized size = 1.54 \begin {gather*} - \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c \sqrt {x}} + \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {3}{2}}} + \frac {4 a b \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{c}} \sqrt {c + d x}} \right )}}{c \sqrt {- \frac {1}{c}}} + b^{2} \left (\begin {cases} \frac {x}{\sqrt {c}} & \text {for}\: d = 0 \\\frac {2 \sqrt {c + d x}}{d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x) + 1)/(c*sqrt(x)) + a**2*d*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/c**(3/2) + 4*a*b*atan(1/
(sqrt(-1/c)*sqrt(c + d*x)))/(c*sqrt(-1/c)) + b**2*Piecewise((x/sqrt(c), Eq(d, 0)), (2*sqrt(c + d*x)/d, True))

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