\(\int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx\) [1770]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 88 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=-\frac {2 (b e-a f)}{f (d e-c f) \sqrt {e+f x}}+\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {79, 65, 214} \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=\frac {2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{3/2}}-\frac {2 (b e-a f)}{f \sqrt {e+f x} (d e-c f)} \]

[In]

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

[Out]

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

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=\frac {2 b e-2 a f}{f (-d e+c f) \sqrt {e+f x}}+\frac {2 (b c-a d) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{\sqrt {d} (-d e+c f)^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90

method result size
pseudoelliptic \(\frac {-\frac {2 \left (a f -b e \right )}{\sqrt {f x +e}}-\frac {2 f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}}{\left (c f -d e \right ) f}\) \(79\)
derivativedivides \(\frac {-\frac {2 \left (a f -b e \right )}{\left (c f -d e \right ) \sqrt {f x +e}}-\frac {2 f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}}{f}\) \(89\)
default \(\frac {-\frac {2 \left (a f -b e \right )}{\left (c f -d e \right ) \sqrt {f x +e}}-\frac {2 f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}}{f}\) \(89\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (76) = 152\).

Time = 0.24 (sec) , antiderivative size = 363, normalized size of antiderivative = 4.12 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=\left [\frac {{\left ({\left (b c - a d\right )} f^{2} x + {\left (b c - a d\right )} e f\right )} \sqrt {d^{2} e - c d f} \log \left (\frac {d f x + 2 \, d e - c f + 2 \, \sqrt {d^{2} e - c d f} \sqrt {f x + e}}{d x + c}\right ) - 2 \, {\left (b d^{2} e^{2} + a c d f^{2} - {\left (b c d + a d^{2}\right )} e f\right )} \sqrt {f x + e}}{d^{3} e^{3} f - 2 \, c d^{2} e^{2} f^{2} + c^{2} d e f^{3} + {\left (d^{3} e^{2} f^{2} - 2 \, c d^{2} e f^{3} + c^{2} d f^{4}\right )} x}, -\frac {2 \, {\left ({\left ({\left (b c - a d\right )} f^{2} x + {\left (b c - a d\right )} e f\right )} \sqrt {-d^{2} e + c d f} \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) + {\left (b d^{2} e^{2} + a c d f^{2} - {\left (b c d + a d^{2}\right )} e f\right )} \sqrt {f x + e}\right )}}{d^{3} e^{3} f - 2 \, c d^{2} e^{2} f^{2} + c^{2} d e f^{3} + {\left (d^{3} e^{2} f^{2} - 2 \, c d^{2} e f^{3} + c^{2} d f^{4}\right )} x}\right ] \]

[In]

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

[Out]

[(((b*c - a*d)*f^2*x + (b*c - a*d)*e*f)*sqrt(d^2*e - c*d*f)*log((d*f*x + 2*d*e - c*f + 2*sqrt(d^2*e - c*d*f)*s
qrt(f*x + e))/(d*x + c)) - 2*(b*d^2*e^2 + a*c*d*f^2 - (b*c*d + a*d^2)*e*f)*sqrt(f*x + e))/(d^3*e^3*f - 2*c*d^2
*e^2*f^2 + c^2*d*e*f^3 + (d^3*e^2*f^2 - 2*c*d^2*e*f^3 + c^2*d*f^4)*x), -2*(((b*c - a*d)*f^2*x + (b*c - a*d)*e*
f)*sqrt(-d^2*e + c*d*f)*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) + (b*d^2*e^2 + a*c*d*f^2 - (b
*c*d + a*d^2)*e*f)*sqrt(f*x + e))/(d^3*e^3*f - 2*c*d^2*e^2*f^2 + c^2*d*e*f^3 + (d^3*e^2*f^2 - 2*c*d^2*e*f^3 +
c^2*d*f^4)*x)]

Sympy [A] (verification not implemented)

Time = 3.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {a f - b e}{\sqrt {e + f x} \left (c f - d e\right )} - \frac {f \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d \sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b x}{d} + \frac {\left (a d - b c\right ) \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d}}{e^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(-(a*f - b*e)/(sqrt(e + f*x)*(c*f - d*e)) - f*(a*d - b*c)*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))
/(d*sqrt((c*f - d*e)/d)*(c*f - d*e)))/f, Ne(f, 0)), ((b*x/d + (a*d - b*c)*Piecewise((x/c, Eq(d, 0)), (log(c +
d*x)/d, True))/d)/e**(3/2), True))

Maxima [F(-2)]

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

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)^(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(c*f-d*e>0)', see `assume?` for
 more detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=-\frac {2 \, {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} {\left (d e - c f\right )}} - \frac {2 \, {\left (b e - a f\right )}}{{\left (d e f - c f^{2}\right )} \sqrt {f x + e}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.09 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{3/2}} \, dx=-\frac {2\,\left (a\,f-b\,e\right )}{f\,\sqrt {e+f\,x}\,\left (c\,f-d\,e\right )}-\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {d}\,\sqrt {e+f\,x}\,\left (a\,d-b\,c\right )}{\left (2\,a\,d-2\,b\,c\right )\,\sqrt {c\,f-d\,e}}\right )\,\left (a\,d-b\,c\right )}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{3/2}} \]

[In]

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

[Out]

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