3.7.22 \(\int \frac {(a+b \sqrt {c+d x})^2}{x} \, dx\) [622]

Optimal. Leaf size=57 \[ b^2 d x+4 a b \sqrt {c+d x}-4 a b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\left (a^2+b^2 c\right ) \log (x) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {378, 1412, 815, 649, 213, 266} \begin {gather*} \log (x) \left (a^2+b^2 c\right )+4 a b \sqrt {c+d x}-4 a b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+b^2 d x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[c + d*x])^2/x,x]

[Out]

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

Rule 213

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 378

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1412

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 63, normalized size = 1.11 \begin {gather*} b \left (b c+b d x+4 a \sqrt {c+d x}\right )-4 a b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\left (a^2+b^2 c\right ) \log (-d x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.30, size = 51, normalized size = 0.89

method	result	size
default	\(b^{2} \left (d x +c \ln \left (x \right )\right )+2 a b \left (2 \sqrt {d x +c}-2 \sqrt {c}\, \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )\right )+a^{2} \ln \left (x \right )\)	\(51\)
derivativedivides	\(\left (d x +c \right ) b^{2}+4 a b \sqrt {d x +c}-\left (-b^{2} c -a^{2}\right ) \ln \left (-d x \right )-4 a b \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) \sqrt {c}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 70, normalized size = 1.23 \begin {gather*} 2 \, a b \sqrt {c} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) + {\left (d x + c\right )} b^{2} + 4 \, \sqrt {d x + c} a b + {\left (b^{2} c + a^{2}\right )} \log \left (d x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 118, normalized size = 2.07 \begin {gather*} \left [b^{2} d x + 2 \, a b \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 4 \, \sqrt {d x + c} a b + {\left (b^{2} c + a^{2}\right )} \log \left (x\right ), b^{2} d x + 4 \, a b \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 4 \, \sqrt {d x + c} a b + {\left (b^{2} c + a^{2}\right )} \log \left (x\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 13.60, size = 65, normalized size = 1.14 \begin {gather*} a^{2} \log {\left (x \right )} - 2 a b \left (- \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - 2 \sqrt {c + d x}\right ) + b^{2} c \log {\left (x \right )} + b^{2} d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) - 2*a*b*(-2*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) - 2*sqrt(c + d*x)) + b**2*c*log(x) + b**2*d*x

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Giac [A]
time = 4.00, size = 59, normalized size = 1.04 \begin {gather*} \frac {4 \, a b c \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + {\left (d x + c\right )} b^{2} + 4 \, \sqrt {d x + c} a b + {\left (b^{2} c + a^{2}\right )} \log \left (d x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.09, size = 130, normalized size = 2.28 \begin {gather*} \ln \left (\left (2\,a^2+2\,c\,b^2\right )\,\sqrt {c+d\,x}-2\,{\left (a+b\,\sqrt {c}\right )}^2\,\sqrt {c+d\,x}+4\,a\,b\,c\right )\,{\left (a+b\,\sqrt {c}\right )}^2+\ln \left (\left (2\,a^2+2\,c\,b^2\right )\,\sqrt {c+d\,x}-2\,{\left (a-b\,\sqrt {c}\right )}^2\,\sqrt {c+d\,x}+4\,a\,b\,c\right )\,{\left (a-b\,\sqrt {c}\right )}^2+4\,a\,b\,\sqrt {c+d\,x}+b^2\,d\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c + d*x)^(1/2))^2/x,x)

[Out]

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

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