3.19 \(\int \frac{\sqrt{a+b x} (e+f x)}{x (c+d x)} \, dx\)

Optimal. Leaf size=101 \[ \frac{2 \sqrt{b c-a d} (d e-c f) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{c d^{3/2}}-\frac{2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{c}+\frac{2 f \sqrt{a+b x}}{d} \]

[Out]

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

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Rubi [A]  time = 0.11897, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {154, 156, 63, 208, 205} \[ \frac{2 \sqrt{b c-a d} (d e-c f) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{c d^{3/2}}-\frac{2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{c}+\frac{2 f \sqrt{a+b x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(e + f*x))/(x*(c + d*x)),x]

[Out]

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

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.218331, size = 100, normalized size = 0.99 \[ \frac{-\frac{2 \sqrt{b c-a d} (c f-d e) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{3/2}}+\frac{2 c f \sqrt{a+b x}}{d}-2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{c} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(e + f*x))/(x*(c + d*x)),x]

[Out]

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

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Maple [A]  time = 0.013, size = 103, normalized size = 1. \begin{align*} 2\,{\frac{f\sqrt{bx+a}}{d}}-2\,{\frac{acdf-a{d}^{2}e-b{c}^{2}f+bcde}{dc\sqrt{ \left ( ad-bc \right ) d}}{\it Artanh} \left ({\frac{\sqrt{bx+a}d}{\sqrt{ \left ( ad-bc \right ) d}}} \right ) }-2\,{\frac{e\sqrt{a}}{c}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64713, size = 1013, normalized size = 10.03 \begin{align*} \left [\frac{\sqrt{a} d e \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \, \sqrt{b x + a} c f -{\left (d e - c f\right )} \sqrt{-\frac{b c - a d}{d}} \log \left (\frac{b d x - b c + 2 \, a d - 2 \, \sqrt{b x + a} d \sqrt{-\frac{b c - a d}{d}}}{d x + c}\right )}{c d}, \frac{2 \, \sqrt{-a} d e \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) + 2 \, \sqrt{b x + a} c f -{\left (d e - c f\right )} \sqrt{-\frac{b c - a d}{d}} \log \left (\frac{b d x - b c + 2 \, a d - 2 \, \sqrt{b x + a} d \sqrt{-\frac{b c - a d}{d}}}{d x + c}\right )}{c d}, \frac{\sqrt{a} d e \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \, \sqrt{b x + a} c f - 2 \,{\left (d e - c f\right )} \sqrt{\frac{b c - a d}{d}} \arctan \left (-\frac{\sqrt{b x + a} d \sqrt{\frac{b c - a d}{d}}}{b c - a d}\right )}{c d}, \frac{2 \,{\left (\sqrt{-a} d e \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) + \sqrt{b x + a} c f -{\left (d e - c f\right )} \sqrt{\frac{b c - a d}{d}} \arctan \left (-\frac{\sqrt{b x + a} d \sqrt{\frac{b c - a d}{d}}}{b c - a d}\right )\right )}}{c d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(a)*d*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*sqrt(b*x + a)*c*f - (d*e - c*f)*sqrt(-(b*c - a*
d)/d)*log((b*d*x - b*c + 2*a*d - 2*sqrt(b*x + a)*d*sqrt(-(b*c - a*d)/d))/(d*x + c)))/(c*d), (2*sqrt(-a)*d*e*ar
ctan(sqrt(b*x + a)*sqrt(-a)/a) + 2*sqrt(b*x + a)*c*f - (d*e - c*f)*sqrt(-(b*c - a*d)/d)*log((b*d*x - b*c + 2*a
*d - 2*sqrt(b*x + a)*d*sqrt(-(b*c - a*d)/d))/(d*x + c)))/(c*d), (sqrt(a)*d*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a
) + 2*a)/x) + 2*sqrt(b*x + a)*c*f - 2*(d*e - c*f)*sqrt((b*c - a*d)/d)*arctan(-sqrt(b*x + a)*d*sqrt((b*c - a*d)
/d)/(b*c - a*d)))/(c*d), 2*(sqrt(-a)*d*e*arctan(sqrt(b*x + a)*sqrt(-a)/a) + sqrt(b*x + a)*c*f - (d*e - c*f)*sq
rt((b*c - a*d)/d)*arctan(-sqrt(b*x + a)*d*sqrt((b*c - a*d)/d)/(b*c - a*d)))/(c*d)]

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Sympy [A]  time = 15.252, size = 100, normalized size = 0.99 \begin{align*} \frac{2 a e \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{c \sqrt{- a}} + \frac{2 f \sqrt{a + b x}}{d} + \frac{2 \left (a d - b c\right ) \left (c f - d e\right ) \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- \frac{a d - b c}{d}}} \right )}}{c d^{2} \sqrt{- \frac{a d - b c}{d}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.62353, size = 151, normalized size = 1.5 \begin{align*} \frac{2 \, a \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) e}{\sqrt{-a} c} + \frac{2 \, \sqrt{b x + a} f}{d} - \frac{2 \,{\left (b c^{2} f - a c d f - b c d e + a d^{2} e\right )} \arctan \left (\frac{\sqrt{b x + a} d}{\sqrt{b c d - a d^{2}}}\right )}{\sqrt{b c d - a d^{2}} c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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