3.1776 \(\int \frac{(a+b x)^2 \sqrt{e+f x}}{c+d x} \, dx\)

Optimal. Leaf size=138 \[ -\frac{2 b (e+f x)^{3/2} (-2 a d f+b c f+b d e)}{3 d^2 f^2}+\frac{2 \sqrt{e+f x} (b c-a d)^2}{d^3}-\frac{2 (b c-a d)^2 \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2}}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2} \]

[Out]

(2*(b*c - a*d)^2*Sqrt[e + f*x])/d^3 - (2*b*(b*d*e + b*c*f - 2*a*d*f)*(e + f*x)^(3/2))/(3*d^2*f^2) + (2*b^2*(e
+ f*x)^(5/2))/(5*d*f^2) - (2*(b*c - a*d)^2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d
^(7/2)

________________________________________________________________________________________

Rubi [A]  time = 0.12469, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {88, 50, 63, 208} \[ -\frac{2 b (e+f x)^{3/2} (-2 a d f+b c f+b d e)}{3 d^2 f^2}+\frac{2 \sqrt{e+f x} (b c-a d)^2}{d^3}-\frac{2 (b c-a d)^2 \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2}}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c - a*d)^2*Sqrt[e + f*x])/d^3 - (2*b*(b*d*e + b*c*f - 2*a*d*f)*(e + f*x)^(3/2))/(3*d^2*f^2) + (2*b^2*(e
+ f*x)^(5/2))/(5*d*f^2) - (2*(b*c - a*d)^2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d
^(7/2)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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)^2 \sqrt{e+f x}}{c+d x} \, dx &=\int \left (-\frac{b (b d e+b c f-2 a d f) \sqrt{e+f x}}{d^2 f}+\frac{(-b c+a d)^2 \sqrt{e+f x}}{d^2 (c+d x)}+\frac{b^2 (e+f x)^{3/2}}{d f}\right ) \, dx\\ &=-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{3 d^2 f^2}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2}+\frac{(b c-a d)^2 \int \frac{\sqrt{e+f x}}{c+d x} \, dx}{d^2}\\ &=\frac{2 (b c-a d)^2 \sqrt{e+f x}}{d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{3 d^2 f^2}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2}+\frac{\left ((b c-a d)^2 (d e-c f)\right ) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{d^3}\\ &=\frac{2 (b c-a d)^2 \sqrt{e+f x}}{d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{3 d^2 f^2}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2}+\frac{\left (2 (b c-a d)^2 (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{d^3 f}\\ &=\frac{2 (b c-a d)^2 \sqrt{e+f x}}{d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{3 d^2 f^2}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2}-\frac{2 (b c-a d)^2 \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.10317, size = 138, normalized size = 1. \[ -\frac{2 b (e+f x)^{3/2} (-2 a d f+b c f+b d e)}{3 d^2 f^2}+\frac{2 \sqrt{e+f x} (b c-a d)^2}{d^3}-\frac{2 (b c-a d)^2 \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2}}+\frac{2 b^2 (e+f x)^{5/2}}{5 d f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c - a*d)^2*Sqrt[e + f*x])/d^3 - (2*b*(b*d*e + b*c*f - 2*a*d*f)*(e + f*x)^(3/2))/(3*d^2*f^2) + (2*b^2*(e
+ f*x)^(5/2))/(5*d*f^2) - (2*(b*c - a*d)^2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d
^(7/2)

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 387, normalized size = 2.8 \begin{align*}{\frac{2\,{b}^{2}}{5\,d{f}^{2}} \left ( fx+e \right ) ^{{\frac{5}{2}}}}+{\frac{4\,ab}{3\,df} \left ( fx+e \right ) ^{{\frac{3}{2}}}}-{\frac{2\,{b}^{2}c}{3\,f{d}^{2}} \left ( fx+e \right ) ^{{\frac{3}{2}}}}-{\frac{2\,{b}^{2}e}{3\,d{f}^{2}} \left ( fx+e \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{a}^{2}\sqrt{fx+e}}{d}}-4\,{\frac{abc\sqrt{fx+e}}{{d}^{2}}}+2\,{\frac{{b}^{2}{c}^{2}\sqrt{fx+e}}{{d}^{3}}}-2\,{\frac{{a}^{2}cf}{d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{{a}^{2}e}{\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+4\,{\frac{ab{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 ) }-4\,{\frac{abce}{d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-2\,{\frac{{b}^{2}{c}^{3}f}{{d}^{3}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{{b}^{2}{c}^{2}e}{{d}^{2}\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)^2*(f*x+e)^(1/2)/(d*x+c),x)

[Out]

2/5*b^2*(f*x+e)^(5/2)/d/f^2+4/3/f/d*(f*x+e)^(3/2)*a*b-2/3/f/d^2*(f*x+e)^(3/2)*b^2*c-2/3/f^2/d*(f*x+e)^(3/2)*b^
2*e+2/d*a^2*(f*x+e)^(1/2)-4/d^2*a*b*c*(f*x+e)^(1/2)+2/d^3*b^2*c^2*(f*x+e)^(1/2)-2*f/d/((c*f-d*e)*d)^(1/2)*arct
an((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*c+2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2)
)*a^2*e+4*f/d^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*c^2-4/d/((c*f-d*e)*d)^(1/2
)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*c*e-2*f/d^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f
-d*e)*d)^(1/2))*b^2*c^3+2/d^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^2*c^2*e

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.40626, size = 863, normalized size = 6.25 \begin{align*} \left [\frac{15 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} \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 (3 \, b^{2} d^{2} f^{2} x^{2} - 2 \, b^{2} d^{2} e^{2} - 5 \,{\left (b^{2} c d - 2 \, a b d^{2}\right )} e f + 15 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} +{\left (b^{2} d^{2} e f - 5 \,{\left (b^{2} c d - 2 \, a b d^{2}\right )} f^{2}\right )} x\right )} \sqrt{f x + e}}{15 \, d^{3} f^{2}}, -\frac{2 \,{\left (15 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} \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 (3 \, b^{2} d^{2} f^{2} x^{2} - 2 \, b^{2} d^{2} e^{2} - 5 \,{\left (b^{2} c d - 2 \, a b d^{2}\right )} e f + 15 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} +{\left (b^{2} d^{2} e f - 5 \,{\left (b^{2} c d - 2 \, a b d^{2}\right )} f^{2}\right )} x\right )} \sqrt{f x + e}\right )}}{15 \, d^{3} f^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*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*(3*b^2*d^2*f^2*x^2 - 2*b^2*d^2*e^2 - 5*(b^2*c*d - 2*a*b*d^2)*e*f + 15*(b^
2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2 + (b^2*d^2*e*f - 5*(b^2*c*d - 2*a*b*d^2)*f^2)*x)*sqrt(f*x + e))/(d^3*f^2), -2
/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/
(d*e - c*f)) - (3*b^2*d^2*f^2*x^2 - 2*b^2*d^2*e^2 - 5*(b^2*c*d - 2*a*b*d^2)*e*f + 15*(b^2*c^2 - 2*a*b*c*d + a^
2*d^2)*f^2 + (b^2*d^2*e*f - 5*(b^2*c*d - 2*a*b*d^2)*f^2)*x)*sqrt(f*x + e))/(d^3*f^2)]

________________________________________________________________________________________

Sympy [A]  time = 7.01214, size = 155, normalized size = 1.12 \begin{align*} \frac{2 \left (\frac{b^{2} \left (e + f x\right )^{\frac{5}{2}}}{5 d f} + \frac{\left (e + f x\right )^{\frac{3}{2}} \left (2 a b d f - b^{2} c f - b^{2} d e\right )}{3 d^{2} f} + \frac{\sqrt{e + f x} \left (a^{2} d^{2} f - 2 a b c d f + b^{2} c^{2} f\right )}{d^{3}} - \frac{f \left (a d - b c\right )^{2} \left (c f - d e\right ) \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{d^{4} \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)**2*(f*x+e)**(1/2)/(d*x+c),x)

[Out]

2*(b**2*(e + f*x)**(5/2)/(5*d*f) + (e + f*x)**(3/2)*(2*a*b*d*f - b**2*c*f - b**2*d*e)/(3*d**2*f) + sqrt(e + f*
x)*(a**2*d**2*f - 2*a*b*c*d*f + b**2*c**2*f)/d**3 - f*(a*d - b*c)**2*(c*f - d*e)*atan(sqrt(e + f*x)/sqrt((c*f
- d*e)/d))/(d**4*sqrt((c*f - d*e)/d)))/f

________________________________________________________________________________________

Giac [B]  time = 1.5432, size = 338, normalized size = 2.45 \begin{align*} -\frac{2 \,{\left (b^{2} c^{3} f - 2 \, a b c^{2} d f + a^{2} c d^{2} f - b^{2} c^{2} d e + 2 \, a b c d^{2} e - a^{2} d^{3} 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^{3}} + \frac{2 \,{\left (3 \,{\left (f x + e\right )}^{\frac{5}{2}} b^{2} d^{4} f^{8} - 5 \,{\left (f x + e\right )}^{\frac{3}{2}} b^{2} c d^{3} f^{9} + 10 \,{\left (f x + e\right )}^{\frac{3}{2}} a b d^{4} f^{9} + 15 \, \sqrt{f x + e} b^{2} c^{2} d^{2} f^{10} - 30 \, \sqrt{f x + e} a b c d^{3} f^{10} + 15 \, \sqrt{f x + e} a^{2} d^{4} f^{10} - 5 \,{\left (f x + e\right )}^{\frac{3}{2}} b^{2} d^{4} f^{8} e\right )}}{15 \, d^{5} f^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^2*c^3*f - 2*a*b*c^2*d*f + a^2*c*d^2*f - b^2*c^2*d*e + 2*a*b*c*d^2*e - a^2*d^3*e)*arctan(sqrt(f*x + e)*d/
sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d^3) + 2/15*(3*(f*x + e)^(5/2)*b^2*d^4*f^8 - 5*(f*x + e)^(3/2)*b^2*c
*d^3*f^9 + 10*(f*x + e)^(3/2)*a*b*d^4*f^9 + 15*sqrt(f*x + e)*b^2*c^2*d^2*f^10 - 30*sqrt(f*x + e)*a*b*c*d^3*f^1
0 + 15*sqrt(f*x + e)*a^2*d^4*f^10 - 5*(f*x + e)^(3/2)*b^2*d^4*f^8*e)/(d^5*f^10)