3.18.85 \(\int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx\) [1785]

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

[Out]

-2/3*b^2*(-3*a*d*f+b*c*f+2*b*d*e)*(f*x+e)^(3/2)/d^2/f^3+2/5*b^3*(f*x+e)^(5/2)/d/f^3+2*(-a*d+b*c)^3*arctanh(d^(
1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(7/2)/(-c*f+d*e)^(1/2)+2*b*(3*a^2*d^2*f^2-3*a*b*d*f*(c*f+d*e)+b^2*(c^2*
f^2+c*d*e*f+d^2*e^2))*(f*x+e)^(1/2)/d^3/f^3

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 184, normalized size of antiderivative = 1.00, 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 = {90, 65, 214} \begin {gather*} \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} \end {gather*}

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 65

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 90

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], 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 ) \text {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 = 0.26, size = 157, normalized size = 0.85 \begin {gather*} \frac {2 b \sqrt {e+f x} \left (45 a^2 d^2 f^2+15 a b d f (-2 d e-3 c f+d f x)+b^2 \left (15 c^2 f^2-5 c d f (-2 e+f x)+d^2 \left (8 e^2-4 e f x+3 f^2 x^2\right )\right )\right )}{15 d^3 f^3}+\frac {2 (-b c+a d)^3 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2} \sqrt {-d e+c f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[e + f*x]*(45*a^2*d^2*f^2 + 15*a*b*d*f*(-2*d*e - 3*c*f + d*f*x) + b^2*(15*c^2*f^2 - 5*c*d*f*(-2*e + f
*x) + d^2*(8*e^2 - 4*e*f*x + 3*f^2*x^2))))/(15*d^3*f^3) + (2*(-(b*c) + a*d)^3*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/S
qrt[-(d*e) + c*f]])/(d^(7/2)*Sqrt[-(d*e) + c*f])

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 249, normalized size = 1.35

method	result	size
derivativedivides	\(\frac {\frac {2 b \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+3 a^{2} d^{2} f^{2} \sqrt {f x +e}-3 a b c d \,f^{2} \sqrt {f x +e}-3 f \,d^{2} e a b \sqrt {f x +e}+b^{2} c^{2} f^{2} \sqrt {f x +e}+b^{2} c d e f \sqrt {f x +e}+b^{2} d^{2} e^{2} \sqrt {f x +e}\right )}{d^{3}}+\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{3}}\)	\(249\)
default	\(\frac {\frac {2 b \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+3 a^{2} d^{2} f^{2} \sqrt {f x +e}-3 a b c d \,f^{2} \sqrt {f x +e}-3 f \,d^{2} e a b \sqrt {f x +e}+b^{2} c^{2} f^{2} \sqrt {f x +e}+b^{2} c d e f \sqrt {f x +e}+b^{2} d^{2} e^{2} \sqrt {f x +e}\right )}{d^{3}}+\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{3}}\)	\(249\)
risch	\(\frac {2 b \left (3 b^{2} d^{2} f^{2} x^{2}+15 a b \,d^{2} f^{2} x -5 b^{2} c d \,f^{2} x -4 b^{2} d^{2} e f x +45 a^{2} d^{2} f^{2}-45 a b c d \,f^{2}-30 a b \,d^{2} e f +15 b^{2} c^{2} f^{2}+10 b^{2} c d e f +8 b^{2} d^{2} e^{2}\right ) \sqrt {f x +e}}{15 f^{3} d^{3}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a^{3}}{\sqrt {\left (c f -d e \right ) d}}-\frac {6 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a^{2} b c}{d \sqrt {\left (c f -d e \right ) d}}+\frac {6 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a \,b^{2} c^{2}}{d^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b^{3} c^{3}}{d^{3} \sqrt {\left (c f -d e \right ) d}}\)	\(300\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*f-%e*d>0)', see `assume?` fo
r more detai

________________________________________________________________________________________

Fricas [A]
time = 1.22, size = 666, normalized size = 3.62 \begin {gather*} \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 {-c d f + d^{2} e} f^{3} \log \left (\frac {d f x - c f + 2 \, d e - 2 \, \sqrt {-c d f + d^{2} e} \sqrt {f x + e}}{d x + c}\right ) + 2 \, {\left (3 \, b^{3} c d^{3} f^{3} x^{2} - 8 \, b^{3} d^{4} e^{3} - 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3}\right )} f^{3} x + 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} + 2 \, {\left (2 \, b^{3} d^{4} f x - {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} f\right )} e^{2} - {\left (3 \, b^{3} d^{4} f^{2} x^{2} - {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} f^{2} x + 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 9 \, a^{2} b d^{4}\right )} f^{2}\right )} e\right )} \sqrt {f x + e}}{15 \, {\left (c d^{4} f^{4} - d^{5} f^{3} e\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 {c d f - d^{2} e} f^{3} \arctan \left (\frac {\sqrt {c d f - d^{2} e} \sqrt {f x + e}}{d f x + d e}\right ) + {\left (3 \, b^{3} c d^{3} f^{3} x^{2} - 8 \, b^{3} d^{4} e^{3} - 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3}\right )} f^{3} x + 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} + 2 \, {\left (2 \, b^{3} d^{4} f x - {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} f\right )} e^{2} - {\left (3 \, b^{3} d^{4} f^{2} x^{2} - {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} f^{2} x + 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 9 \, a^{2} b d^{4}\right )} f^{2}\right )} e\right )} \sqrt {f x + e}\right )}}{15 \, {\left (c d^{4} f^{4} - d^{5} f^{3} e\right )}}\right ] \end {gather*}

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

________________________________________________________________________________________

Sympy [A]
time = 31.76, size = 201, normalized size = 1.09 \begin {gather*} \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}} \cdot \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 {gather*}

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))

________________________________________________________________________________________

Giac [A]
time = 0.63, size = 298, normalized size = 1.62 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

Mupad [B]
time = 0.10, size = 264, normalized size = 1.43 \begin {gather*} \sqrt {e+f\,x}\,\left (\frac {\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{d^2\,f^6}\right )\,\left (c\,f^4-d\,e\,f^3\right )}{d\,f^3}+\frac {6\,b\,{\left (a\,f-b\,e\right )}^2}{d\,f^3}\right )-{\left (e+f\,x\right )}^{3/2}\,\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{3\,d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{3\,d^2\,f^6}\right )+\frac {2\,b^3\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f^3}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3}{\sqrt {c\,f-d\,e}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{d^{7/2}\,\sqrt {c\,f-d\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e + f*x)^(1/2)*((((6*b^3*e - 6*a*b^2*f)/(d*f^3) + (2*b^3*(c*f^4 - d*e*f^3))/(d^2*f^6))*(c*f^4 - d*e*f^3))/(d*
f^3) + (6*b*(a*f - b*e)^2)/(d*f^3)) - (e + f*x)^(3/2)*((6*b^3*e - 6*a*b^2*f)/(3*d*f^3) + (2*b^3*(c*f^4 - d*e*f
^3))/(3*d^2*f^6)) + (2*b^3*(e + f*x)^(5/2))/(5*d*f^3) + (2*atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^3)/((c*f
- d*e)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)))*(a*d - b*c)^3)/(d^(7/2)*(c*f - d*e)^(1/2))

________________________________________________________________________________________