\(\int \frac {A+B x}{(d+e x) (b x+c x^2)^{3/2}} \, dx\) [1203]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (A b (c d-b e)+c (2 A c d-b (B d+A e)) x)}{b^2 d (c d-b e) \sqrt {b x+c x^2}}-\frac {e (B d-A e) \text {arctanh}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{d^{3/2} (c d-b e)^{3/2}} \]

[Out]

-e*(-A*e+B*d)*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1/2)/(c*x^2+b*x)^(1/2))/d^(3/2)/(-b*e+c*d)^
(3/2)-2*(A*b*(-b*e+c*d)+c*(2*A*c*d-b*(A*e+B*d))*x)/b^2/d/(-b*e+c*d)/(c*x^2+b*x)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {836, 12, 738, 212} \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {e (B d-A e) \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}}-\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (c d-b e)} \]

[In]

Int[(A + B*x)/((d + e*x)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-2*(A*b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x))/(b^2*d*(c*d - b*e)*Sqrt[b*x + c*x^2]) - (e*(B*d - A*e)*
ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(d^(3/2)*(c*d - b*e)^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 836

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (A b (c d-b e)+c (2 A c d-b (B d+A e)) x)}{b^2 d (c d-b e) \sqrt {b x+c x^2}}-\frac {2 \int \frac {b^2 e (B d-A e)}{2 (d+e x) \sqrt {b x+c x^2}} \, dx}{b^2 d (c d-b e)} \\ & = -\frac {2 (A b (c d-b e)+c (2 A c d-b (B d+A e)) x)}{b^2 d (c d-b e) \sqrt {b x+c x^2}}-\frac {(e (B d-A e)) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{d (c d-b e)} \\ & = -\frac {2 (A b (c d-b e)+c (2 A c d-b (B d+A e)) x)}{b^2 d (c d-b e) \sqrt {b x+c x^2}}+\frac {(2 e (B d-A e)) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac {-b d-(2 c d-b e) x}{\sqrt {b x+c x^2}}\right )}{d (c d-b e)} \\ & = -\frac {2 (A b (c d-b e)+c (2 A c d-b (B d+A e)) x)}{b^2 d (c d-b e) \sqrt {b x+c x^2}}-\frac {e (B d-A e) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{d^{3/2} (c d-b e)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (\sqrt {d} \sqrt {-c d+b e} \left (b B c d x+A \left (-b c d+b^2 e-2 c^2 d x+b c e x\right )\right )+b^2 e (B d-A e) \sqrt {x} \sqrt {b+c x} \arctan \left (\frac {-e \sqrt {x} \sqrt {b+c x}+\sqrt {c} (d+e x)}{\sqrt {d} \sqrt {-c d+b e}}\right )\right )}{b^2 d^{3/2} (-c d+b e)^{3/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[(A + B*x)/((d + e*x)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-2*(Sqrt[d]*Sqrt[-(c*d) + b*e]*(b*B*c*d*x + A*(-(b*c*d) + b^2*e - 2*c^2*d*x + b*c*e*x)) + b^2*e*(B*d - A*e)*S
qrt[x]*Sqrt[b + c*x]*ArcTan[(-(e*Sqrt[x]*Sqrt[b + c*x]) + Sqrt[c]*(d + e*x))/(Sqrt[d]*Sqrt[-(c*d) + b*e])]))/(
b^2*d^(3/2)*(-(c*d) + b*e)^(3/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(\frac {-\frac {2 A \sqrt {x \left (c x +b \right )}}{d x}+\frac {2 \left (A e -B d \right ) b^{2} e \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )}{\sqrt {d \left (b e -c d \right )}\, d \left (b e -c d \right )}+\frac {2 c \left (A c -B b \right ) x}{\left (b e -c d \right ) \sqrt {x \left (c x +b \right )}}}{b^{2}}\)	\(120\)
risch	\(-\frac {2 A \left (c x +b \right )}{b^{2} d \sqrt {x \left (c x +b \right )}}-\frac {-\frac {b \left (A e -B d \right ) \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}-\frac {2 \left (A c -B b \right ) d \sqrt {c \left (x +\frac {b}{c}\right )^{2}-b \left (x +\frac {b}{c}\right )}}{\left (b e -c d \right ) b \left (x +\frac {b}{c}\right )}}{b d}\)	\(238\)
default	\(-\frac {2 B \left (2 c x +b \right )}{e \,b^{2} \sqrt {c \,x^{2}+b x}}+\frac {\left (A e -B d \right ) \left (-\frac {e^{2}}{d \left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{d \left (b e -c d \right ) \left (-\frac {4 c d \left (b e -c d \right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {e^{2} \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}\right )}{e^{2}}\)	\(375\)

[In]

int((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*(-A/d*(x*(c*x+b))^(1/2)/x+(A*e-B*d)*b^2*e/(d*(b*e-c*d))^(1/2)*arctan((x*(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1/
2))/d/(b*e-c*d)+c*(A*c-B*b)/(b*e-c*d)*x/(x*(c*x+b))^(1/2))/b^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (127) = 254\).

Time = 0.40 (sec) , antiderivative size = 543, normalized size of antiderivative = 3.85 \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {\sqrt {c d^{2} - b d e} {\left ({\left (B b^{2} c d e - A b^{2} c e^{2}\right )} x^{2} + {\left (B b^{3} d e - A b^{3} e^{2}\right )} x\right )} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) - 2 \, {\left (A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} + {\left (A b^{2} c d e^{2} - {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{3} + {\left (B b^{2} c - 3 \, A b c^{2}\right )} d^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}, -\frac {2 \, {\left (\sqrt {-c d^{2} + b d e} {\left ({\left (B b^{2} c d e - A b^{2} c e^{2}\right )} x^{2} + {\left (B b^{3} d e - A b^{3} e^{2}\right )} x\right )} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) + {\left (A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} + {\left (A b^{2} c d e^{2} - {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{3} + {\left (B b^{2} c - 3 \, A b c^{2}\right )} d^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}\right ] \]

[In]

integrate((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[(sqrt(c*d^2 - b*d*e)*((B*b^2*c*d*e - A*b^2*c*e^2)*x^2 + (B*b^3*d*e - A*b^3*e^2)*x)*log((b*d + (2*c*d - b*e)*x
- 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(A*b*c^2*d^3 - 2*A*b^2*c*d^2*e + A*b^3*d*e^2 + (A*b
^2*c*d*e^2 - (B*b*c^2 - 2*A*c^3)*d^3 + (B*b^2*c - 3*A*b*c^2)*d^2*e)*x)*sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^
3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2*d^4 - 2*b^4*c*d^3*e + b^5*d^2*e^2)*x), -2*(sqrt(-c*d^2 + b*d*e)*((
B*b^2*c*d*e - A*b^2*c*e^2)*x^2 + (B*b^3*d*e - A*b^3*e^2)*x)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c
*d - b*e)*x)) + (A*b*c^2*d^3 - 2*A*b^2*c*d^2*e + A*b^3*d*e^2 + (A*b^2*c*d*e^2 - (B*b*c^2 - 2*A*c^3)*d^3 + (B*b
^2*c - 3*A*b*c^2)*d^2*e)*x)*sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2
*d^4 - 2*b^4*c*d^3*e + b^5*d^2*e^2)*x)]

Sympy [F]

\[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \]

[In]

integrate((B*x+A)/(e*x+d)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral((A + B*x)/((x*(b + c*x))**(3/2)*(d + e*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),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((b/e-(2*c*d)/e^2)^2>0)', see `
assume?` for

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (B b c d^{2} - 2 \, A c^{2} d^{2} + A b c d e\right )} x}{b^{2} c d^{3} - b^{3} d^{2} e} - \frac {A b c d^{2} - A b^{2} d e}{b^{2} c d^{3} - b^{3} d^{2} e}\right )}}{\sqrt {c x^{2} + b x}} + \frac {2 \, {\left (B d e - A e^{2}\right )} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right )}{{\left (c d^{2} - b d e\right )} \sqrt {-c d^{2} + b d e}} \]

[In]

integrate((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

2*((B*b*c*d^2 - 2*A*c^2*d^2 + A*b*c*d*e)*x/(b^2*c*d^3 - b^3*d^2*e) - (A*b*c*d^2 - A*b^2*d*e)/(b^2*c*d^3 - b^3*
d^2*e))/sqrt(c*x^2 + b*x) + 2*(B*d*e - A*e^2)*arctan(((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d
^2 + b*d*e))/((c*d^2 - b*d*e)*sqrt(-c*d^2 + b*d*e))

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(d+e x) \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \]

[In]

int((A + B*x)/((b*x + c*x^2)^(3/2)*(d + e*x)),x)

[Out]

int((A + B*x)/((b*x + c*x^2)^(3/2)*(d + e*x)), x)