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

Optimal. Leaf size=77 \[ -\frac{2 (a+b x)^{3/2} (2 a d f-5 b (c f+d e)-3 b d f x)}{15 b^2}+2 c e \sqrt{a+b x}-2 \sqrt{a} c e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right ) \]

[Out]

2*c*e*Sqrt[a + b*x] - (2*(a + b*x)^(3/2)*(2*a*d*f - 5*b*(d*e + c*f) - 3*b*d*f*x))/(15*b^2) - 2*Sqrt[a]*c*e*Arc
Tanh[Sqrt[a + b*x]/Sqrt[a]]

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Rubi [A]  time = 0.0235912, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {147, 50, 63, 208} \[ -\frac{2 (a+b x)^{3/2} (2 a d f-5 b (c f+d e)-3 b d f x)}{15 b^2}+2 c e \sqrt{a+b x}-2 \sqrt{a} c e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*c*e*Sqrt[a + b*x] - (2*(a + b*x)^(3/2)*(2*a*d*f - 5*b*(d*e + c*f) - 3*b*d*f*x))/(15*b^2) - 2*Sqrt[a]*c*e*Arc
Tanh[Sqrt[a + b*x]/Sqrt[a]]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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{\sqrt{a+b x} (c+d x) (e+f x)}{x} \, dx &=-\frac{2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}+(c e) \int \frac{\sqrt{a+b x}}{x} \, dx\\ &=2 c e \sqrt{a+b x}-\frac{2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}+(a c e) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=2 c e \sqrt{a+b x}-\frac{2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}+\frac{(2 a c e) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{b}\\ &=2 c e \sqrt{a+b x}-\frac{2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}-2 \sqrt{a} c e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.123062, size = 87, normalized size = 1.13 \[ \frac{2 (a+b x)^{3/2} (-a d f+b c f+b d e)}{3 b^2}+\frac{2 d f (a+b x)^{5/2}}{5 b^2}+2 c e \sqrt{a+b x}-2 \sqrt{a} c e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 89, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{{b}^{2}} \left ( 1/5\,df \left ( bx+a \right ) ^{5/2}-1/3\, \left ( bx+a \right ) ^{3/2}adf+1/3\, \left ( bx+a \right ) ^{3/2}bcf+1/3\, \left ( bx+a \right ) ^{3/2}bde+{b}^{2}ce\sqrt{bx+a}-\sqrt{a}{b}^{2}ce{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^2*(1/5*d*f*(b*x+a)^(5/2)-1/3*(b*x+a)^(3/2)*a*d*f+1/3*(b*x+a)^(3/2)*b*c*f+1/3*(b*x+a)^(3/2)*b*d*e+b^2*c*e*(
b*x+a)^(1/2)-a^(1/2)*b^2*c*e*arctanh((b*x+a)^(1/2)/a^(1/2)))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.22554, size = 517, normalized size = 6.71 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{2} c e \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (3 \, b^{2} d f x^{2} + 5 \,{\left (3 \, b^{2} c + a b d\right )} e +{\left (5 \, a b c - 2 \, a^{2} d\right )} f +{\left (5 \, b^{2} d e +{\left (5 \, b^{2} c + a b d\right )} f\right )} x\right )} \sqrt{b x + a}}{15 \, b^{2}}, \frac{2 \,{\left (15 \, \sqrt{-a} b^{2} c e \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (3 \, b^{2} d f x^{2} + 5 \,{\left (3 \, b^{2} c + a b d\right )} e +{\left (5 \, a b c - 2 \, a^{2} d\right )} f +{\left (5 \, b^{2} d e +{\left (5 \, b^{2} c + a b d\right )} f\right )} x\right )} \sqrt{b x + a}\right )}}{15 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 18.1863, size = 92, normalized size = 1.19 \begin{align*} \frac{2 a c e \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + 2 c e \sqrt{a + b x} + \frac{2 d f \left (a + b x\right )^{\frac{5}{2}}}{5 b^{2}} + \frac{2 \left (a + b x\right )^{\frac{3}{2}} \left (- a d f + b c f + b d e\right )}{3 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.75584, size = 142, normalized size = 1.84 \begin{align*} \frac{2 \, a c \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) e}{\sqrt{-a}} + \frac{2 \,{\left (5 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{9} c f + 3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{8} d f - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{8} d f + 15 \, \sqrt{b x + a} b^{10} c e + 5 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{9} d e\right )}}{15 \, b^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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