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

Optimal. Leaf size=122 \[ -\frac {2 \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}}+\frac {2 \left (a g^2-b f g+c f^2\right )}{g^2 \sqrt {f+g x} (e f-d g)}+\frac {2 c \sqrt {f+g x}}{e g^2} \]

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Rubi [A]  time = 0.22, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {897, 1261, 208} \begin {gather*} -\frac {2 \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}}+\frac {2 \left (a g^2-b f g+c f^2\right )}{g^2 \sqrt {f+g x} (e f-d g)}+\frac {2 c \sqrt {f+g x}}{e g^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*f^2 - b*f*g + a*g^2))/(g^2*(e*f - d*g)*Sqrt[f + g*x]) + (2*c*Sqrt[f + g*x])/(e*g^2) - (2*(c*d^2 - b*d*e
+ a*e^2)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(e^(3/2)*(e*f - d*g)^(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 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\frac {c f^2-b f g+a g^2}{g^2}-\frac {(2 c f-b g) x^2}{g^2}+\frac {c x^4}{g^2}}{x^2 \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {c}{e g}+\frac {c f^2-b f g+a g^2}{g (-e f+d g) x^2}-\frac {\left (c d^2-b d e+a e^2\right ) g}{e (e f-d g) \left (e f-d g-e x^2\right )}\right ) \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=\frac {2 \left (c f^2-b f g+a g^2\right )}{g^2 (e f-d g) \sqrt {f+g x}}+\frac {2 c \sqrt {f+g x}}{e g^2}-\frac {\left (2 \left (c d^2-b d e+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{e f-d g-e x^2} \, dx,x,\sqrt {f+g x}\right )}{e (e f-d g)}\\ &=\frac {2 \left (c f^2-b f g+a g^2\right )}{g^2 (e f-d g) \sqrt {f+g x}}+\frac {2 c \sqrt {f+g x}}{e g^2}-\frac {2 \left (c d^2-b d e+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 124, normalized size = 1.02 \begin {gather*} \frac {2 \left (-\frac {g^2 \left (c d^2-e (b d-a e)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}}+\frac {c f^2-g (b f-a g)}{\sqrt {f+g x} (e f-d g)}+\frac {c \sqrt {f+g x}}{e}\right )}{g^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 139, normalized size = 1.14 \begin {gather*} \frac {2 \left (a e^2-b d e+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x} \sqrt {d g-e f}}{e f-d g}\right )}{e^{3/2} (d g-e f)^{3/2}}+\frac {2 \left (a e g^2-b e f g-c d g (f+g x)+c e f^2+c e f (f+g x)\right )}{e g^2 \sqrt {f+g x} (e f-d g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 540, normalized size = 4.43 \begin {gather*} \left [-\frac {{\left ({\left (c d^{2} - b d e + a e^{2}\right )} g^{3} x + {\left (c d^{2} - b d e + a e^{2}\right )} f g^{2}\right )} \sqrt {e^{2} f - d e g} \log \left (\frac {e g x + 2 \, e f - d g + 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) - 2 \, {\left (2 \, c e^{3} f^{3} - a d e^{2} g^{3} - {\left (3 \, c d e^{2} + b e^{3}\right )} f^{2} g + {\left (c d^{2} e + b d e^{2} + a e^{3}\right )} f g^{2} + {\left (c e^{3} f^{2} g - 2 \, c d e^{2} f g^{2} + c d^{2} e g^{3}\right )} x\right )} \sqrt {g x + f}}{e^{4} f^{3} g^{2} - 2 \, d e^{3} f^{2} g^{3} + d^{2} e^{2} f g^{4} + {\left (e^{4} f^{2} g^{3} - 2 \, d e^{3} f g^{4} + d^{2} e^{2} g^{5}\right )} x}, \frac {2 \, {\left ({\left ({\left (c d^{2} - b d e + a e^{2}\right )} g^{3} x + {\left (c d^{2} - b d e + a e^{2}\right )} f g^{2}\right )} \sqrt {-e^{2} f + d e g} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) + {\left (2 \, c e^{3} f^{3} - a d e^{2} g^{3} - {\left (3 \, c d e^{2} + b e^{3}\right )} f^{2} g + {\left (c d^{2} e + b d e^{2} + a e^{3}\right )} f g^{2} + {\left (c e^{3} f^{2} g - 2 \, c d e^{2} f g^{2} + c d^{2} e g^{3}\right )} x\right )} \sqrt {g x + f}\right )}}{e^{4} f^{3} g^{2} - 2 \, d e^{3} f^{2} g^{3} + d^{2} e^{2} f g^{4} + {\left (e^{4} f^{2} g^{3} - 2 \, d e^{3} f g^{4} + d^{2} e^{2} g^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 112, normalized size = 0.92 \begin {gather*} -\frac {2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{{\left (d g e - f e^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, \sqrt {g x + f} c e^{\left (-1\right )}}{g^{2}} - \frac {2 \, {\left (c f^{2} - b f g + a g^{2}\right )}}{{\left (d g^{3} - f g^{2} e\right )} \sqrt {g x + f}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 237, normalized size = 1.94 \begin {gather*} -\frac {2 a e \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}+\frac {2 b d \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 c \,d^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}\, e}-\frac {2 a}{\left (d g -e f \right ) \sqrt {g x +f}}+\frac {2 b f}{\left (d g -e f \right ) \sqrt {g x +f}\, g}-\frac {2 c \,f^{2}}{\left (d g -e f \right ) \sqrt {g x +f}\, g^{2}}+\frac {2 \sqrt {g x +f}\, c}{e \,g^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((c*x^2+b*x+a)/(e*x+d)/(g*x+f)^(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(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [B]  time = 3.21, size = 162, normalized size = 1.33 \begin {gather*} \frac {2\,c\,\sqrt {f+g\,x}}{e\,g^2}+\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {f+g\,x}\,\left (e^2\,f-d\,e\,g\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{\sqrt {e}\,{\left (d\,g-e\,f\right )}^{3/2}\,\left (2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2\right )}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{e^{3/2}\,{\left (d\,g-e\,f\right )}^{3/2}}-\frac {2\,\left (c\,e\,f^2-b\,e\,f\,g+a\,e\,g^2\right )}{e\,g^2\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 52.23, size = 116, normalized size = 0.95 \begin {gather*} \frac {2 c \sqrt {f + g x}}{e g^{2}} - \frac {2 \left (a g^{2} - b f g + c f^{2}\right )}{g^{2} \sqrt {f + g x} \left (d g - e f\right )} - \frac {2 \left (a e^{2} - b d e + c d^{2}\right ) \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{2} \sqrt {\frac {d g - e f}{e}} \left (d g - e f\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*sqrt(f + g*x)/(e*g**2) - 2*(a*g**2 - b*f*g + c*f**2)/(g**2*sqrt(f + g*x)*(d*g - e*f)) - 2*(a*e**2 - b*d*e
+ c*d**2)*atan(sqrt(f + g*x)/sqrt((d*g - e*f)/e))/(e**2*sqrt((d*g - e*f)/e)*(d*g - e*f))

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