3.2272 \(\int \frac {\sqrt {d+e x} (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac {2 \sqrt {d+e x} (-b e g+c d g+c e f)}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {788, 660, 208} \[ \frac {2 \sqrt {d+e x} (-b e g+c d g+c e f)}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 660

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

Rule 788

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 147, normalized size = 0.95 \[ \frac {2 \sqrt {d+e x} \left (c \sqrt {2 c d-b e} (d g-e f) \sqrt {c (d-e x)-b e} \tanh ^{-1}\left (\frac {\sqrt {-b e+c d-c e x}}{\sqrt {2 c d-b e}}\right )+(2 c d-b e) (-b e g+c d g+c e f)\right )}{c e^2 (b e-2 c d)^2 \sqrt {(d+e x) (c (d-e x)-b e)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.76, size = 776, normalized size = 5.01 \[ \left [-\frac {{\left ({\left (c^{2} e^{3} f - c^{2} d e^{2} g\right )} x^{2} - {\left (c^{2} d^{2} e - b c d e^{2}\right )} f + {\left (c^{2} d^{3} - b c d^{2} e\right )} g + {\left (b c e^{3} f - b c d e^{2} g\right )} x\right )} \sqrt {2 \, c d - b e} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} x + 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {2 \, c d - b e} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left ({\left (2 \, c^{2} d e - b c e^{2}\right )} f + {\left (2 \, c^{2} d^{2} - 3 \, b c d e + b^{2} e^{2}\right )} g\right )} \sqrt {e x + d}}{4 \, c^{4} d^{4} e^{2} - 8 \, b c^{3} d^{3} e^{3} + 5 \, b^{2} c^{2} d^{2} e^{4} - b^{3} c d e^{5} - {\left (4 \, c^{4} d^{2} e^{4} - 4 \, b c^{3} d e^{5} + b^{2} c^{2} e^{6}\right )} x^{2} - {\left (4 \, b c^{3} d^{2} e^{4} - 4 \, b^{2} c^{2} d e^{5} + b^{3} c e^{6}\right )} x}, \frac {2 \, {\left ({\left ({\left (c^{2} e^{3} f - c^{2} d e^{2} g\right )} x^{2} - {\left (c^{2} d^{2} e - b c d e^{2}\right )} f + {\left (c^{2} d^{3} - b c d^{2} e\right )} g + {\left (b c e^{3} f - b c d e^{2} g\right )} x\right )} \sqrt {-2 \, c d + b e} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {-2 \, c d + b e} \sqrt {e x + d}}{c e^{2} x^{2} + b e^{2} x - c d^{2} + b d e}\right ) + \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left ({\left (2 \, c^{2} d e - b c e^{2}\right )} f + {\left (2 \, c^{2} d^{2} - 3 \, b c d e + b^{2} e^{2}\right )} g\right )} \sqrt {e x + d}\right )}}{4 \, c^{4} d^{4} e^{2} - 8 \, b c^{3} d^{3} e^{3} + 5 \, b^{2} c^{2} d^{2} e^{4} - b^{3} c d e^{5} - {\left (4 \, c^{4} d^{2} e^{4} - 4 \, b c^{3} d e^{5} + b^{2} c^{2} e^{6}\right )} x^{2} - {\left (4 \, b c^{3} d^{2} e^{4} - 4 \, b^{2} c^{2} d e^{5} + b^{3} c e^{6}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(((c^2*e^3*f - c^2*d*e^2*g)*x^2 - (c^2*d^2*e - b*c*d*e^2)*f + (c^2*d^3 - b*c*d^2*e)*g + (b*c*e^3*f - b*c*d*e
^2*g)*x)*sqrt(2*c*d - b*e)*log(-(c*e^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x + 2*sqrt(-c*e^2*x^2 - b*e
^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*sqrt(-c*e^2*x^2 - b*e^2*
x + c*d^2 - b*d*e)*((2*c^2*d*e - b*c*e^2)*f + (2*c^2*d^2 - 3*b*c*d*e + b^2*e^2)*g)*sqrt(e*x + d))/(4*c^4*d^4*e
^2 - 8*b*c^3*d^3*e^3 + 5*b^2*c^2*d^2*e^4 - b^3*c*d*e^5 - (4*c^4*d^2*e^4 - 4*b*c^3*d*e^5 + b^2*c^2*e^6)*x^2 - (
4*b*c^3*d^2*e^4 - 4*b^2*c^2*d*e^5 + b^3*c*e^6)*x), 2*(((c^2*e^3*f - c^2*d*e^2*g)*x^2 - (c^2*d^2*e - b*c*d*e^2)
*f + (c^2*d^3 - b*c*d^2*e)*g + (b*c*e^3*f - b*c*d*e^2*g)*x)*sqrt(-2*c*d + b*e)*arctan(sqrt(-c*e^2*x^2 - b*e^2*
x + c*d^2 - b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(c*e^2*x^2 + b*e^2*x - c*d^2 + b*d*e)) + sqrt(-c*e^2*x^2 -
b*e^2*x + c*d^2 - b*d*e)*((2*c^2*d*e - b*c*e^2)*f + (2*c^2*d^2 - 3*b*c*d*e + b^2*e^2)*g)*sqrt(e*x + d))/(4*c^
4*d^4*e^2 - 8*b*c^3*d^3*e^3 + 5*b^2*c^2*d^2*e^4 - b^3*c*d*e^5 - (4*c^4*d^2*e^4 - 4*b*c^3*d*e^5 + b^2*c^2*e^6)*
x^2 - (4*b*c^3*d^2*e^4 - 4*b^2*c^2*d*e^5 + b^3*c*e^6)*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [b,c,d,exp(1),exp(2)]=[72,91,-18,-31,46]Warning, need to choose a branch for the root of a polyn
omial with parameters. This might be wrong.The choice was done assuming [b,c,d,exp(1),exp(2)]=[-76,-17,63,68,9
8]Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choic
e was done assuming [b,c,d,exp(1),exp(2)]=[-95,-95,-57,19,-77]Warning, need to choose a branch for the root of
a polynomial with parameters. This might be wrong.The choice was done assuming [b,c,d,exp(1),exp(2)]=[-29,-98
,-9,8,96]Evaluation time: 111.17Unable to transpose Error: Bad Argument Value

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maple [A]  time = 0.07, size = 207, normalized size = 1.34 \[ -\frac {2 \left (\sqrt {-c e x -b e +c d}\, c d g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-\sqrt {-c e x -b e +c d}\, c e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+\sqrt {b e -2 c d}\, b e g -\sqrt {b e -2 c d}\, c d g -\sqrt {b e -2 c d}\, c e f \right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}{\left (b e -2 c d \right )^{\frac {3}{2}} \left (c e x +b e -c d \right ) \sqrt {e x +d}\, c \,e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e x + d} {\left (g x + f\right )}}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(e*x + d)*(g*x + f)/(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (f+g\,x\right )\,\sqrt {d+e\,x}}{{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d + e x} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d + e*x)*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(3/2), x)

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