3.1768 \(\int \frac{(a+b x) \sqrt{e+f x}}{c+d x} \, dx\)

Optimal. Leaf size=98 \[ -\frac{2 \sqrt{e+f x} (b c-a d)}{d^2}+\frac{2 (b c-a d) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2}}+\frac{2 b (e+f x)^{3/2}}{3 d f} \]

[Out]

(-2*(b*c - a*d)*Sqrt[e + f*x])/d^2 + (2*b*(e + f*x)^(3/2))/(3*d*f) + (2*(b*c - a*d)*Sqrt[d*e - c*f]*ArcTanh[(S
qrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(5/2)

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Rubi [A]  time = 0.0790366, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \[ -\frac{2 \sqrt{e+f x} (b c-a d)}{d^2}+\frac{2 (b c-a d) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2}}+\frac{2 b (e+f x)^{3/2}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c - a*d)*Sqrt[e + f*x])/d^2 + (2*b*(e + f*x)^(3/2))/(3*d*f) + (2*(b*c - a*d)*Sqrt[d*e - c*f]*ArcTanh[(S
qrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(5/2)

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 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 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) \sqrt{e+f x}}{c+d x} \, dx &=\frac{2 b (e+f x)^{3/2}}{3 d f}+\frac{\left (2 \left (-\frac{3}{2} b c f+\frac{3 a d f}{2}\right )\right ) \int \frac{\sqrt{e+f x}}{c+d x} \, dx}{3 d f}\\ &=-\frac{2 (b c-a d) \sqrt{e+f x}}{d^2}+\frac{2 b (e+f x)^{3/2}}{3 d f}-\frac{((b c-a d) (d e-c f)) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{d^2}\\ &=-\frac{2 (b c-a d) \sqrt{e+f x}}{d^2}+\frac{2 b (e+f x)^{3/2}}{3 d f}-\frac{(2 (b c-a d) (d e-c f)) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{d^2 f}\\ &=-\frac{2 (b c-a d) \sqrt{e+f x}}{d^2}+\frac{2 b (e+f x)^{3/2}}{3 d f}+\frac{2 (b c-a d) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.109859, size = 94, normalized size = 0.96 \[ \frac{2 \sqrt{e+f x} (3 a d f-3 b c f+b d (e+f x))}{3 d^2 f}+\frac{2 (b c-a d) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e + f*x]*(-3*b*c*f + 3*a*d*f + b*d*(e + f*x)))/(3*d^2*f) + (2*(b*c - a*d)*Sqrt[d*e - c*f]*ArcTanh[(Sqr
t[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(5/2)

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Maple [B]  time = 0.009, size = 211, normalized size = 2.2 \begin{align*}{\frac{2\,b}{3\,df} \left ( fx+e \right ) ^{{\frac{3}{2}}}}+2\,{\frac{a\sqrt{fx+e}}{d}}-2\,{\frac{bc\sqrt{fx+e}}{{d}^{2}}}-2\,{\frac{acf}{d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{ae}{\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{b{c}^{2}f}{{d}^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-2\,{\frac{bce}{d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b*(f*x+e)^(3/2)/d/f+2/d*a*(f*x+e)^(1/2)-2/d^2*b*c*(f*x+e)^(1/2)-2*f/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(
1/2)*d/((c*f-d*e)*d)^(1/2))*a*c+2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*e+2*f/d^2/
((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b*c^2-2/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(
1/2)*d/((c*f-d*e)*d)^(1/2))*b*c*e

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38578, size = 471, normalized size = 4.81 \begin{align*} \left [-\frac{3 \,{\left (b c - a d\right )} f \sqrt{\frac{d e - c f}{d}} \log \left (\frac{d f x + 2 \, d e - c f - 2 \, \sqrt{f x + e} d \sqrt{\frac{d e - c f}{d}}}{d x + c}\right ) - 2 \,{\left (b d f x + b d e - 3 \,{\left (b c - a d\right )} f\right )} \sqrt{f x + e}}{3 \, d^{2} f}, \frac{2 \,{\left (3 \,{\left (b c - a d\right )} f \sqrt{-\frac{d e - c f}{d}} \arctan \left (-\frac{\sqrt{f x + e} d \sqrt{-\frac{d e - c f}{d}}}{d e - c f}\right ) +{\left (b d f x + b d e - 3 \,{\left (b c - a d\right )} f\right )} \sqrt{f x + e}\right )}}{3 \, d^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*(b*c - a*d)*f*sqrt((d*e - c*f)/d)*log((d*f*x + 2*d*e - c*f - 2*sqrt(f*x + e)*d*sqrt((d*e - c*f)/d))/(
d*x + c)) - 2*(b*d*f*x + b*d*e - 3*(b*c - a*d)*f)*sqrt(f*x + e))/(d^2*f), 2/3*(3*(b*c - a*d)*f*sqrt(-(d*e - c*
f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e - c*f)) + (b*d*f*x + b*d*e - 3*(b*c - a*d)*f)*sqrt(f*x
+ e))/(d^2*f)]

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Sympy [A]  time = 6.16722, size = 94, normalized size = 0.96 \begin{align*} \frac{2 \left (\frac{b \left (e + f x\right )^{\frac{3}{2}}}{3 d} + \frac{\sqrt{e + f x} \left (a d f - b c f\right )}{d^{2}} - \frac{f \left (a d - b c\right ) \left (c f - d e\right ) \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{d^{3} \sqrt{\frac{c f - d e}{d}}}\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*(e + f*x)**(3/2)/(3*d) + sqrt(e + f*x)*(a*d*f - b*c*f)/d**2 - f*(a*d - b*c)*(c*f - d*e)*atan(sqrt(e + f*x
)/sqrt((c*f - d*e)/d))/(d**3*sqrt((c*f - d*e)/d)))/f

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Giac [A]  time = 1.79462, size = 176, normalized size = 1.8 \begin{align*} \frac{2 \,{\left (b c^{2} f - a c d f - b c d e + a d^{2} e\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{\sqrt{c d f - d^{2} e} d^{2}} + \frac{2 \,{\left ({\left (f x + e\right )}^{\frac{3}{2}} b d^{2} f^{2} - 3 \, \sqrt{f x + e} b c d f^{3} + 3 \, \sqrt{f x + e} a d^{2} f^{3}\right )}}{3 \, d^{3} f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*c^2*f - a*c*d*f - b*c*d*e + a*d^2*e)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d^2
) + 2/3*((f*x + e)^(3/2)*b*d^2*f^2 - 3*sqrt(f*x + e)*b*c*d*f^3 + 3*sqrt(f*x + e)*a*d^2*f^3)/(d^3*f^3)