3.1785 \(\int \frac{(a+b x)^3}{(c+d x) \sqrt{e+f x}} \, dx\)

Optimal. Leaf size=184 \[ \frac{2 b \sqrt{e+f x} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{d^3 f^3}-\frac{2 b^2 (e+f x)^{3/2} (-3 a d f+b c f+2 b d e)}{3 d^2 f^3}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2} \sqrt{d e-c f}}+\frac{2 b^3 (e+f x)^{5/2}}{5 d f^3} \]

[Out]

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

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Rubi [A]  time = 0.172326, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {88, 63, 208} \[ \frac{2 b \sqrt{e+f x} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{d^3 f^3}-\frac{2 b^2 (e+f x)^{3/2} (-3 a d f+b c f+2 b d e)}{3 d^2 f^3}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2} \sqrt{d e-c f}}+\frac{2 b^3 (e+f x)^{5/2}}{5 d f^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/((c + d*x)*Sqrt[e + f*x]),x]

[Out]

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

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

Mathematica [A]  time = 0.142687, size = 184, normalized size = 1. \[ \frac{2 b \sqrt{e+f x} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{d^3 f^3}-\frac{2 b^2 (e+f x)^{3/2} (-3 a d f+b c f+2 b d e)}{3 d^2 f^3}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{7/2} \sqrt{d e-c f}}+\frac{2 b^3 (e+f x)^{5/2}}{5 d f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3/((c + d*x)*Sqrt[e + f*x]),x]

[Out]

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

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Maple [B]  time = 0.011, size = 372, normalized size = 2. \begin{align*}{\frac{2\,{b}^{3}}{5\,d{f}^{3}} \left ( fx+e \right ) ^{{\frac{5}{2}}}}+2\,{\frac{{b}^{2} \left ( fx+e \right ) ^{3/2}a}{d{f}^{2}}}-{\frac{2\,{b}^{3}c}{3\,{d}^{2}{f}^{2}} \left ( fx+e \right ) ^{{\frac{3}{2}}}}-{\frac{4\,{b}^{3}e}{3\,d{f}^{3}} \left ( fx+e \right ) ^{{\frac{3}{2}}}}+6\,{\frac{{a}^{2}b\sqrt{fx+e}}{df}}-6\,{\frac{a{b}^{2}c\sqrt{fx+e}}{{d}^{2}f}}-6\,{\frac{a{b}^{2}e\sqrt{fx+e}}{d{f}^{2}}}+2\,{\frac{{b}^{3}{c}^{2}\sqrt{fx+e}}{f{d}^{3}}}+2\,{\frac{ce{b}^{3}\sqrt{fx+e}}{{d}^{2}{f}^{2}}}+2\,{\frac{{b}^{3}{e}^{2}\sqrt{fx+e}}{d{f}^{3}}}+2\,{\frac{{a}^{3}}{\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-6\,{\frac{{a}^{2}bc}{d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+6\,{\frac{a{b}^{2}{c}^{2}}{{d}^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-2\,{\frac{{b}^{3}{c}^{3}}{{d}^{3}\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)^3/(d*x+c)/(f*x+e)^(1/2),x)

[Out]

2/5*b^3*(f*x+e)^(5/2)/d/f^3+2/f^2*b^2/d*(f*x+e)^(3/2)*a-2/3/f^2*b^3/d^2*(f*x+e)^(3/2)*c-4/3/f^3*b^3/d*(f*x+e)^
(3/2)*e+6/f*b/d*a^2*(f*x+e)^(1/2)-6/f*b^2/d^2*a*c*(f*x+e)^(1/2)-6/f^2*b^2/d*a*e*(f*x+e)^(1/2)+2/f*b^3/d^3*c^2*
(f*x+e)^(1/2)+2/f^2*b^3/d^2*c*e*(f*x+e)^(1/2)+2/f^3*b^3/d*e^2*(f*x+e)^(1/2)+2/((c*f-d*e)*d)^(1/2)*arctan((f*x+
e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^3-6/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*c*
b+6/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^2*c^2-2/d^3/((c*f-d*e)*d)^(1/2)*ar
ctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^3*c^3

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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)^3/(d*x+c)/(f*x+e)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36327, size = 1335, normalized size = 7.26 \begin{align*} \left [-\frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{d^{2} e - c d f} f^{3} \log \left (\frac{d f x + 2 \, d e - c f - 2 \, \sqrt{d^{2} e - c d f} \sqrt{f x + e}}{d x + c}\right ) - 2 \,{\left (8 \, b^{3} d^{4} e^{3} + 2 \,{\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e^{2} f + 5 \,{\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 9 \, a^{2} b d^{4}\right )} e f^{2} - 15 \,{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3}\right )} f^{3} + 3 \,{\left (b^{3} d^{4} e f^{2} - b^{3} c d^{3} f^{3}\right )} x^{2} -{\left (4 \, b^{3} d^{4} e^{2} f +{\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e f^{2} - 5 \,{\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3}\right )} f^{3}\right )} x\right )} \sqrt{f x + e}}{15 \,{\left (d^{5} e f^{3} - c d^{4} f^{4}\right )}}, -\frac{2 \,{\left (15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-d^{2} e + c d f} f^{3} \arctan \left (\frac{\sqrt{-d^{2} e + c d f} \sqrt{f x + e}}{d f x + d e}\right ) -{\left (8 \, b^{3} d^{4} e^{3} + 2 \,{\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e^{2} f + 5 \,{\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 9 \, a^{2} b d^{4}\right )} e f^{2} - 15 \,{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3}\right )} f^{3} + 3 \,{\left (b^{3} d^{4} e f^{2} - b^{3} c d^{3} f^{3}\right )} x^{2} -{\left (4 \, b^{3} d^{4} e^{2} f +{\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e f^{2} - 5 \,{\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3}\right )} f^{3}\right )} x\right )} \sqrt{f x + e}\right )}}{15 \,{\left (d^{5} e f^{3} - c d^{4} f^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

[-1/15*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(d^2*e - c*d*f)*f^3*log((d*f*x + 2*d*e - c*
f - 2*sqrt(d^2*e - c*d*f)*sqrt(f*x + e))/(d*x + c)) - 2*(8*b^3*d^4*e^3 + 2*(b^3*c*d^3 - 15*a*b^2*d^4)*e^2*f +
5*(b^3*c^2*d^2 - 3*a*b^2*c*d^3 + 9*a^2*b*d^4)*e*f^2 - 15*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3)*f^3 + 3
*(b^3*d^4*e*f^2 - b^3*c*d^3*f^3)*x^2 - (4*b^3*d^4*e^2*f + (b^3*c*d^3 - 15*a*b^2*d^4)*e*f^2 - 5*(b^3*c^2*d^2 -
3*a*b^2*c*d^3)*f^3)*x)*sqrt(f*x + e))/(d^5*e*f^3 - c*d^4*f^4), -2/15*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*
d^2 - a^3*d^3)*sqrt(-d^2*e + c*d*f)*f^3*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) - (8*b^3*d^4*
e^3 + 2*(b^3*c*d^3 - 15*a*b^2*d^4)*e^2*f + 5*(b^3*c^2*d^2 - 3*a*b^2*c*d^3 + 9*a^2*b*d^4)*e*f^2 - 15*(b^3*c^3*d
- 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3)*f^3 + 3*(b^3*d^4*e*f^2 - b^3*c*d^3*f^3)*x^2 - (4*b^3*d^4*e^2*f + (b^3*c*d^
3 - 15*a*b^2*d^4)*e*f^2 - 5*(b^3*c^2*d^2 - 3*a*b^2*c*d^3)*f^3)*x)*sqrt(f*x + e))/(d^5*e*f^3 - c*d^4*f^4)]

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Sympy [A]  time = 43.4855, size = 201, normalized size = 1.09 \begin{align*} \frac{2 b^{3} \left (e + f x\right )^{\frac{5}{2}}}{5 d f^{3}} + \frac{2 b^{2} \left (e + f x\right )^{\frac{3}{2}} \left (3 a d f - b c f - 2 b d e\right )}{3 d^{2} f^{3}} + \frac{2 b \sqrt{e + f x} \left (3 a^{2} d^{2} f^{2} - 3 a b c d f^{2} - 3 a b d^{2} e f + b^{2} c^{2} f^{2} + b^{2} c d e f + b^{2} d^{2} e^{2}\right )}{d^{3} f^{3}} - \frac{2 \left (a d - b c\right )^{3} \operatorname{atan}{\left (\frac{1}{\sqrt{\frac{d}{c f - d e}} \sqrt{e + f x}} \right )}}{d^{3} \sqrt{\frac{d}{c f - d e}} \left (c f - d e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/(d*x+c)/(f*x+e)**(1/2),x)

[Out]

2*b**3*(e + f*x)**(5/2)/(5*d*f**3) + 2*b**2*(e + f*x)**(3/2)*(3*a*d*f - b*c*f - 2*b*d*e)/(3*d**2*f**3) + 2*b*s
qrt(e + f*x)*(3*a**2*d**2*f**2 - 3*a*b*c*d*f**2 - 3*a*b*d**2*e*f + b**2*c**2*f**2 + b**2*c*d*e*f + b**2*d**2*e
**2)/(d**3*f**3) - 2*(a*d - b*c)**3*atan(1/(sqrt(d/(c*f - d*e))*sqrt(e + f*x)))/(d**3*sqrt(d/(c*f - d*e))*(c*f
- d*e))

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Giac [A]  time = 2.19978, size = 402, normalized size = 2.18 \begin{align*} -\frac{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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^{3} d^{4} f^{12} - 5 \,{\left (f x + e\right )}^{\frac{3}{2}} b^{3} c d^{3} f^{13} + 15 \,{\left (f x + e\right )}^{\frac{3}{2}} a b^{2} d^{4} f^{13} + 15 \, \sqrt{f x + e} b^{3} c^{2} d^{2} f^{14} - 45 \, \sqrt{f x + e} a b^{2} c d^{3} f^{14} + 45 \, \sqrt{f x + e} a^{2} b d^{4} f^{14} - 10 \,{\left (f x + e\right )}^{\frac{3}{2}} b^{3} d^{4} f^{12} e + 15 \, \sqrt{f x + e} b^{3} c d^{3} f^{13} e - 45 \, \sqrt{f x + e} a b^{2} d^{4} f^{13} e + 15 \, \sqrt{f x + e} b^{3} d^{4} f^{12} e^{2}\right )}}{15 \, d^{5} f^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

-2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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^3*d^4*f^12 - 5*(f*x + e)^(3/2)*b^3*c*d^3*f^13 + 15*(f*x + e)^(3/2)*
a*b^2*d^4*f^13 + 15*sqrt(f*x + e)*b^3*c^2*d^2*f^14 - 45*sqrt(f*x + e)*a*b^2*c*d^3*f^14 + 45*sqrt(f*x + e)*a^2*
b*d^4*f^14 - 10*(f*x + e)^(3/2)*b^3*d^4*f^12*e + 15*sqrt(f*x + e)*b^3*c*d^3*f^13*e - 45*sqrt(f*x + e)*a*b^2*d^
4*f^13*e + 15*sqrt(f*x + e)*b^3*d^4*f^12*e^2)/(d^5*f^15)