3.6.100 \(\int \frac {a+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx\) [600]

Optimal. Leaf size=112 \[ \frac {2 \left (c f^2+a g^2\right )}{g^2 (e f-d g) \sqrt {f+g x}}+\frac {2 c \sqrt {f+g x}}{e g^2}-\frac {2 \left (c d^2+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 112, 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 = {912, 1275, 214} \begin {gather*} -\frac {2 \left (a e^2+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+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 + c*x^2)/((d + e*x)*(f + g*x)^(3/2)),x]

[Out]

(2*(c*f^2 + 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 + 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 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]

Rule 912

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

Rule 1275

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+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx &=\frac {2 \text {Subst}\left (\int \frac {\frac {c f^2+a g^2}{g^2}-\frac {2 c f 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 \text {Subst}\left (\int \left (\frac {c}{e g}+\frac {c f^2+a g^2}{g (-e f+d g) x^2}-\frac {\left (c d^2+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+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+a e^2\right )\right ) \text {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+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+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 = 0.33, size = 114, normalized size = 1.02 \begin {gather*} -\frac {2 \left (a e g^2-c d g (f+g x)+c e f (2 f+g x)\right )}{e g^2 (-e f+d g) \sqrt {f+g x}}-\frac {2 \left (c d^2+a e^2\right ) \tan ^{-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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 112, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\frac {2 c \sqrt {g x +f}}{e}-\frac {2 \left (a \,g^{2}+c \,f^{2}\right )}{\left (d g -e f \right ) \sqrt {g x +f}}-\frac {2 g^{2} \left (a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) e \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\)	\(112\)
default	\(\frac {\frac {2 c \sqrt {g x +f}}{e}-\frac {2 \left (a \,g^{2}+c \,f^{2}\right )}{\left (d g -e f \right ) \sqrt {g x +f}}-\frac {2 g^{2} \left (a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) e \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\)	\(112\)
risch	\(\frac {2 c \sqrt {g x +f}}{e \,g^{2}}-\frac {2 a}{\left (d g -e f \right ) \sqrt {g x +f}}-\frac {2 c \,f^{2}}{g^{2} \left (d g -e f \right ) \sqrt {g x +f}}-\frac {2 e \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) a}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) c \,d^{2}}{e \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+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(4*%e^2*f-4*%e*d*g>0)', see `as
sume?` for m

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (97) = 194\).
time = 3.33, size = 475, normalized size = 4.24 \begin {gather*} \left [\frac {{\left (c d^{2} g^{3} x + c d^{2} f g^{2} + {\left (a g^{3} x + a f g^{2}\right )} e^{2}\right )} \sqrt {-d g e + f e^{2}} \log \left (-\frac {d g - {\left (g x + 2 \, f\right )} e + 2 \, \sqrt {-d g e + f e^{2}} \sqrt {g x + f}}{x e + d}\right ) + 2 \, \sqrt {g x + f} {\left ({\left (c f^{2} g x + 2 \, c f^{3} + a f g^{2}\right )} e^{3} - {\left (2 \, c d f g^{2} x + 3 \, c d f^{2} g + a d g^{3}\right )} e^{2} + {\left (c d^{2} g^{3} x + c d^{2} f g^{2}\right )} e\right )}}{{\left (f^{2} g^{3} x + f^{3} g^{2}\right )} e^{4} - 2 \, {\left (d f g^{4} x + d f^{2} g^{3}\right )} e^{3} + {\left (d^{2} g^{5} x + d^{2} f g^{4}\right )} e^{2}}, \frac {2 \, {\left ({\left (c d^{2} g^{3} x + c d^{2} f g^{2} + {\left (a g^{3} x + a f g^{2}\right )} e^{2}\right )} \sqrt {d g e - f e^{2}} \arctan \left (-\frac {\sqrt {d g e - f e^{2}} \sqrt {g x + f}}{d g - f e}\right ) + \sqrt {g x + f} {\left ({\left (c f^{2} g x + 2 \, c f^{3} + a f g^{2}\right )} e^{3} - {\left (2 \, c d f g^{2} x + 3 \, c d f^{2} g + a d g^{3}\right )} e^{2} + {\left (c d^{2} g^{3} x + c d^{2} f g^{2}\right )} e\right )}\right )}}{{\left (f^{2} g^{3} x + f^{3} g^{2}\right )} e^{4} - 2 \, {\left (d f g^{4} x + d f^{2} g^{3}\right )} e^{3} + {\left (d^{2} g^{5} x + d^{2} f g^{4}\right )} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 13.72, size = 104, normalized size = 0.93 \begin {gather*} \frac {2 c \sqrt {f + g x}}{e g^{2}} - \frac {2 \left (a g^{2} + c f^{2}\right )}{g^{2} \sqrt {f + g x} \left (d g - e f\right )} - \frac {2 \left (a e^{2} + 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+a)/(e*x+d)/(g*x+f)**(3/2),x)

[Out]

2*c*sqrt(f + g*x)/(e*g**2) - 2*(a*g**2 + c*f**2)/(g**2*sqrt(f + g*x)*(d*g - e*f)) - 2*(a*e**2 + c*d**2)*atan(s
qrt(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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Giac [A]
time = 1.09, size = 101, normalized size = 0.90 \begin {gather*} -\frac {2 \, {\left (c d^{2} + 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} + 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+a)/(e*x+d)/(g*x+f)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.14, size = 141, normalized size = 1.26 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {f+g\,x}\,\left (c\,d^2+a\,e^2\right )\,\left (e^2\,f-d\,e\,g\right )}{\sqrt {e}\,\left (2\,c\,d^2+2\,a\,e^2\right )\,{\left (d\,g-e\,f\right )}^{3/2}}\right )\,\left (c\,d^2+a\,e^2\right )}{e^{3/2}\,{\left (d\,g-e\,f\right )}^{3/2}}+\frac {2\,c\,\sqrt {f+g\,x}}{e\,g^2}-\frac {2\,\left (c\,e\,f^2+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 + c*x^2)/((f + g*x)^(3/2)*(d + e*x)),x)

[Out]

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

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