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

Optimal. Leaf size=145 \[ \frac{2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (2 c f+d e)+5 b^2 c (2 c f+7 d e)\right )+3 b d x (-4 a d f+4 b c f+7 b d e)\right )}{105 b^3}+2 c^2 e \sqrt{a+b x}-2 \sqrt{a} c^2 e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b} \]

[Out]

2*c^2*e*Sqrt[a + b*x] + (2*f*(a + b*x)^(3/2)*(c + d*x)^2)/(7*b) + (2*(a + b*x)^(3/2)*(2*(4*a^2*d^2*f - 7*a*b*d
*(d*e + 2*c*f) + 5*b^2*c*(7*d*e + 2*c*f)) + 3*b*d*(7*b*d*e + 4*b*c*f - 4*a*d*f)*x))/(105*b^3) - 2*Sqrt[a]*c^2*
e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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Rubi [A]  time = 0.0938083, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {153, 147, 50, 63, 208} \[ \frac{2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (2 c f+d e)+5 b^2 c (2 c f+7 d e)\right )+3 b d x (-4 a d f+4 b c f+7 b d e)\right )}{105 b^3}+2 c^2 e \sqrt{a+b x}-2 \sqrt{a} c^2 e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*c^2*e*Sqrt[a + b*x] + (2*f*(a + b*x)^(3/2)*(c + d*x)^2)/(7*b) + (2*(a + b*x)^(3/2)*(2*(4*a^2*d^2*f - 7*a*b*d
*(d*e + 2*c*f) + 5*b^2*c*(7*d*e + 2*c*f)) + 3*b*d*(7*b*d*e + 4*b*c*f - 4*a*d*f)*x))/(105*b^3) - 2*Sqrt[a]*c^2*
e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

Rule 153

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] && IntegerQ[m]

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)^2 (e+f x)}{x} \, dx &=\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac{2 \int \frac{\sqrt{a+b x} (c+d x) \left (\frac{7 b c e}{2}+\frac{1}{2} (7 b d e+4 b c f-4 a d f) x\right )}{x} \, dx}{7 b}\\ &=\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac{2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}+\left (c^2 e\right ) \int \frac{\sqrt{a+b x}}{x} \, dx\\ &=2 c^2 e \sqrt{a+b x}+\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac{2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}+\left (a c^2 e\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=2 c^2 e \sqrt{a+b x}+\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac{2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}+\frac{\left (2 a c^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{b}\\ &=2 c^2 e \sqrt{a+b x}+\frac{2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac{2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}-2 \sqrt{a} c^2 e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.162919, size = 146, normalized size = 1.01 \[ \frac{2 \left (7 b e \left (15 b^2 c^2 \sqrt{a+b x}-15 \sqrt{a} b^2 c^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+5 d (a+b x)^{3/2} (2 b c-a d)+3 d^2 (a+b x)^{5/2}\right )+f (a+b x)^{3/2} \left (42 d (a+b x) (b c-a d)+35 (b c-a d)^2+15 d^2 (a+b x)^2\right )\right )}{105 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(f*(a + b*x)^(3/2)*(35*(b*c - a*d)^2 + 42*d*(b*c - a*d)*(a + b*x) + 15*d^2*(a + b*x)^2) + 7*b*e*(15*b^2*c^2
*Sqrt[a + b*x] + 5*d*(2*b*c - a*d)*(a + b*x)^(3/2) + 3*d^2*(a + b*x)^(5/2) - 15*Sqrt[a]*b^2*c^2*ArcTanh[Sqrt[a
+ b*x]/Sqrt[a]])))/(105*b^3)

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Maple [A]  time = 0.007, size = 176, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{{b}^{3}} \left ( 1/7\,{d}^{2}f \left ( bx+a \right ) ^{7/2}-2/5\, \left ( bx+a \right ) ^{5/2}a{d}^{2}f+2/5\, \left ( bx+a \right ) ^{5/2}bcdf+1/5\, \left ( bx+a \right ) ^{5/2}b{d}^{2}e+1/3\, \left ( bx+a \right ) ^{3/2}{a}^{2}{d}^{2}f-2/3\, \left ( bx+a \right ) ^{3/2}abcdf-1/3\, \left ( bx+a \right ) ^{3/2}ab{d}^{2}e+1/3\, \left ( bx+a \right ) ^{3/2}{b}^{2}{c}^{2}f+2/3\, \left ( bx+a \right ) ^{3/2}{b}^{2}cde+{b}^{3}{c}^{2}e\sqrt{bx+a}-\sqrt{a}{b}^{3}{c}^{2}e{\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)^2*(f*x+e)*(b*x+a)^(1/2)/x,x)

[Out]

2/b^3*(1/7*d^2*f*(b*x+a)^(7/2)-2/5*(b*x+a)^(5/2)*a*d^2*f+2/5*(b*x+a)^(5/2)*b*c*d*f+1/5*(b*x+a)^(5/2)*b*d^2*e+1
/3*(b*x+a)^(3/2)*a^2*d^2*f-2/3*(b*x+a)^(3/2)*a*b*c*d*f-1/3*(b*x+a)^(3/2)*a*b*d^2*e+1/3*(b*x+a)^(3/2)*b^2*c^2*f
+2/3*(b*x+a)^(3/2)*b^2*c*d*e+b^3*c^2*e*(b*x+a)^(1/2)-a^(1/2)*b^3*c^2*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)^2*(f*x+e)*(b*x+a)^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36354, size = 914, normalized size = 6.3 \begin{align*} \left [\frac{105 \, \sqrt{a} b^{3} c^{2} e \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (15 \, b^{3} d^{2} f x^{3} + 3 \,{\left (7 \, b^{3} d^{2} e +{\left (14 \, b^{3} c d + a b^{2} d^{2}\right )} f\right )} x^{2} + 7 \,{\left (15 \, b^{3} c^{2} + 10 \, a b^{2} c d - 2 \, a^{2} b d^{2}\right )} e +{\left (35 \, a b^{2} c^{2} - 28 \, a^{2} b c d + 8 \, a^{3} d^{2}\right )} f +{\left (7 \,{\left (10 \, b^{3} c d + a b^{2} d^{2}\right )} e +{\left (35 \, b^{3} c^{2} + 14 \, a b^{2} c d - 4 \, a^{2} b d^{2}\right )} f\right )} x\right )} \sqrt{b x + a}}{105 \, b^{3}}, \frac{2 \,{\left (105 \, \sqrt{-a} b^{3} c^{2} e \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, b^{3} d^{2} f x^{3} + 3 \,{\left (7 \, b^{3} d^{2} e +{\left (14 \, b^{3} c d + a b^{2} d^{2}\right )} f\right )} x^{2} + 7 \,{\left (15 \, b^{3} c^{2} + 10 \, a b^{2} c d - 2 \, a^{2} b d^{2}\right )} e +{\left (35 \, a b^{2} c^{2} - 28 \, a^{2} b c d + 8 \, a^{3} d^{2}\right )} f +{\left (7 \,{\left (10 \, b^{3} c d + a b^{2} d^{2}\right )} e +{\left (35 \, b^{3} c^{2} + 14 \, a b^{2} c d - 4 \, a^{2} b d^{2}\right )} f\right )} x\right )} \sqrt{b x + a}\right )}}{105 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*sqrt(a)*b^3*c^2*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*b^3*d^2*f*x^3 + 3*(7*b^3*d^
2*e + (14*b^3*c*d + a*b^2*d^2)*f)*x^2 + 7*(15*b^3*c^2 + 10*a*b^2*c*d - 2*a^2*b*d^2)*e + (35*a*b^2*c^2 - 28*a^2
*b*c*d + 8*a^3*d^2)*f + (7*(10*b^3*c*d + a*b^2*d^2)*e + (35*b^3*c^2 + 14*a*b^2*c*d - 4*a^2*b*d^2)*f)*x)*sqrt(b
*x + a))/b^3, 2/105*(105*sqrt(-a)*b^3*c^2*e*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*b^3*d^2*f*x^3 + 3*(7*b^3*d^
2*e + (14*b^3*c*d + a*b^2*d^2)*f)*x^2 + 7*(15*b^3*c^2 + 10*a*b^2*c*d - 2*a^2*b*d^2)*e + (35*a*b^2*c^2 - 28*a^2
*b*c*d + 8*a^3*d^2)*f + (7*(10*b^3*c*d + a*b^2*d^2)*e + (35*b^3*c^2 + 14*a*b^2*c*d - 4*a^2*b*d^2)*f)*x)*sqrt(b
*x + a))/b^3]

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Sympy [A]  time = 18.6783, size = 167, normalized size = 1.15 \begin{align*} \frac{2 a c^{2} e \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + 2 c^{2} e \sqrt{a + b x} + \frac{2 d^{2} f \left (a + b x\right )^{\frac{7}{2}}}{7 b^{3}} + \frac{2 \left (a + b x\right )^{\frac{5}{2}} \left (- 2 a d^{2} f + 2 b c d f + b d^{2} e\right )}{5 b^{3}} + \frac{2 \left (a + b x\right )^{\frac{3}{2}} \left (a^{2} d^{2} f - 2 a b c d f - a b d^{2} e + b^{2} c^{2} f + 2 b^{2} c d e\right )}{3 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*c**2*e*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*c**2*e*sqrt(a + b*x) + 2*d**2*f*(a + b*x)**(7/2)/(7*b**3)
+ 2*(a + b*x)**(5/2)*(-2*a*d**2*f + 2*b*c*d*f + b*d**2*e)/(5*b**3) + 2*(a + b*x)**(3/2)*(a**2*d**2*f - 2*a*b*
c*d*f - a*b*d**2*e + b**2*c**2*f + 2*b**2*c*d*e)/(3*b**3)

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Giac [A]  time = 1.75434, size = 271, normalized size = 1.87 \begin{align*} \frac{2 \, a c^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) e}{\sqrt{-a}} + \frac{2 \,{\left (35 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{20} c^{2} f + 42 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{19} c d f - 70 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{19} c d f + 15 \,{\left (b x + a\right )}^{\frac{7}{2}} b^{18} d^{2} f - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a b^{18} d^{2} f + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2} b^{18} d^{2} f + 105 \, \sqrt{b x + a} b^{21} c^{2} e + 70 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{20} c d e + 21 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{19} d^{2} e - 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{19} d^{2} e\right )}}{105 \, b^{21}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*c^2*arctan(sqrt(b*x + a)/sqrt(-a))*e/sqrt(-a) + 2/105*(35*(b*x + a)^(3/2)*b^20*c^2*f + 42*(b*x + a)^(5/2)*
b^19*c*d*f - 70*(b*x + a)^(3/2)*a*b^19*c*d*f + 15*(b*x + a)^(7/2)*b^18*d^2*f - 42*(b*x + a)^(5/2)*a*b^18*d^2*f
+ 35*(b*x + a)^(3/2)*a^2*b^18*d^2*f + 105*sqrt(b*x + a)*b^21*c^2*e + 70*(b*x + a)^(3/2)*b^20*c*d*e + 21*(b*x
+ a)^(5/2)*b^19*d^2*e - 35*(b*x + a)^(3/2)*a*b^19*d^2*e)/b^21