3.7.1 \(\int \frac {a+c x^2}{(d+e x)^2 (f+g x)^{3/2}} \, dx\) [601]

Optimal. Leaf size=144 \[ -\frac {2 \left (c f^2+a g^2\right )}{g (e f-d g)^2 \sqrt {f+g x}}-\frac {\left (c d^2+a e^2\right ) \sqrt {f+g x}}{e (e f-d g)^2 (d+e x)}+\frac {\left (3 a e^2 g+c d (4 e f-d g)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{5/2}} \]

[Out]

(3*a*e^2*g+c*d*(-d*g+4*e*f))*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(3/2)/(-d*g+e*f)^(5/2)-2*(a*g^2
+c*f^2)/g/(-d*g+e*f)^2/(g*x+f)^(1/2)-(a*e^2+c*d^2)*(g*x+f)^(1/2)/e/(-d*g+e*f)^2/(e*x+d)

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Rubi [A]
time = 0.18, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {912, 1273, 464, 214} \begin {gather*} -\frac {\sqrt {f+g x} \left (a e^2+c d^2\right )}{e (d+e x) (e f-d g)^2}+\frac {\left (3 a e^2 g+c d (4 e f-d g)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{5/2}}-\frac {2 \left (a g^2+c f^2\right )}{g \sqrt {f+g x} (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c*f^2 + a*g^2))/(g*(e*f - d*g)^2*Sqrt[f + g*x]) - ((c*d^2 + a*e^2)*Sqrt[f + g*x])/(e*(e*f - d*g)^2*(d + e
*x)) + ((3*a*e^2*g + c*d*(4*e*f - d*g))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(e^(3/2)*(e*f - d*g)
^(5/2))

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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rubi steps

\begin {align*} \int \frac {a+c x^2}{(d+e x)^2 (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 )^2} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=-\frac {\left (c d^2+a e^2\right ) \sqrt {f+g x}}{e (e f-d g)^2 (d+e x)}-\frac {g^3 \text {Subst}\left (\int \frac {\frac {2 e^2 (e f-d g) \left (c f^2+a g^2\right )}{g^5}+\frac {e \left (a e^2 g^2-c \left (2 e^2 f^2-4 d e f g+d^2 g^2\right )\right ) x^2}{g^5}}{x^2 \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )} \, dx,x,\sqrt {f+g x}\right )}{e^2 (e f-d g)^2}\\ &=-\frac {2 \left (c f^2+a g^2\right )}{g (e f-d g)^2 \sqrt {f+g x}}-\frac {\left (c d^2+a e^2\right ) \sqrt {f+g x}}{e (e f-d g)^2 (d+e x)}-\frac {\left (3 a e^2 g+c d (4 e f-d g)\right ) \text {Subst}\left (\int \frac {1}{\frac {-e f+d g}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{e g (e f-d g)^2}\\ &=-\frac {2 \left (c f^2+a g^2\right )}{g (e f-d g)^2 \sqrt {f+g x}}-\frac {\left (c d^2+a e^2\right ) \sqrt {f+g x}}{e (e f-d g)^2 (d+e x)}+\frac {\left (3 a e^2 g+c d (4 e f-d g)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 148, normalized size = 1.03 \begin {gather*} \frac {-c \left (2 d e f^2+2 e^2 f^2 x+d^2 g (f+g x)\right )-a e g (2 d g+e (f+3 g x))}{e g (e f-d g)^2 (d+e x) \sqrt {f+g x}}+\frac {\left (-3 a e^2 g+c d (-4 e f+d g)\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{3/2} (-e f+d g)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(c*(2*d*e*f^2 + 2*e^2*f^2*x + d^2*g*(f + g*x))) - a*e*g*(2*d*g + e*(f + 3*g*x)))/(e*g*(e*f - d*g)^2*(d + e*x
)*Sqrt[f + g*x]) + ((-3*a*e^2*g + c*d*(-4*e*f + d*g))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/(e^(
3/2)*(-(e*f) + d*g)^(5/2))

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Maple [A]
time = 0.07, size = 152, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {-\frac {2 g \left (\frac {g \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {g x +f}}{2 e \left (e \left (g x +f \right )+d g -e f \right )}+\frac {\left (3 a \,e^{2} g -c \,d^{2} g +4 c d e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 e \sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{2}}-\frac {2 \left (a \,g^{2}+c \,f^{2}\right )}{\left (d g -e f \right )^{2} \sqrt {g x +f}}}{g}\)	\(152\)
default	\(\frac {-\frac {2 g \left (\frac {g \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {g x +f}}{2 e \left (e \left (g x +f \right )+d g -e f \right )}+\frac {\left (3 a \,e^{2} g -c \,d^{2} g +4 c d e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 e \sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{2}}-\frac {2 \left (a \,g^{2}+c \,f^{2}\right )}{\left (d g -e f \right )^{2} \sqrt {g x +f}}}{g}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/g*(-g/(d*g-e*f)^2*(1/2*g*(a*e^2+c*d^2)/e*(g*x+f)^(1/2)/(e*(g*x+f)+d*g-e*f)+1/2*(3*a*e^2*g-c*d^2*g+4*c*d*e*f)
/e/((d*g-e*f)*e)^(1/2)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)))-(a*g^2+c*f^2)/(d*g-e*f)^2/(g*x+f)^(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)^2/(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. 437 vs. \(2 (131) = 262\).
time = 3.04, size = 890, normalized size = 6.18 \begin {gather*} \left [-\frac {{\left (c d^{3} g^{3} x + c d^{3} f g^{2} - 3 \, {\left (a g^{3} x^{2} + a f g^{2} x\right )} e^{3} - {\left (4 \, c d f g^{2} x^{2} + 3 \, a d f g^{2} + {\left (4 \, c d f^{2} g + 3 \, a d g^{3}\right )} x\right )} e^{2} + {\left (c d^{2} g^{3} x^{2} - 3 \, c d^{2} f g^{2} x - 4 \, c d^{2} f^{2} g\right )} e\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 (a f^{2} g + {\left (2 \, c f^{3} + 3 \, a f g^{2}\right )} x\right )} e^{4} + {\left (2 \, c d f^{3} + a d f g^{2} - {\left (2 \, c d f^{2} g + 3 \, a d g^{3}\right )} x\right )} e^{3} + {\left (c d^{2} f g^{2} x - c d^{2} f^{2} g - 2 \, a d^{2} g^{3}\right )} e^{2} - {\left (c d^{3} g^{3} x + c d^{3} f g^{2}\right )} e\right )}}{2 \, {\left ({\left (f^{3} g^{2} x^{2} + f^{4} g x\right )} e^{6} - {\left (3 \, d f^{2} g^{3} x^{2} + 2 \, d f^{3} g^{2} x - d f^{4} g\right )} e^{5} + 3 \, {\left (d^{2} f g^{4} x^{2} - d^{2} f^{3} g^{2}\right )} e^{4} - {\left (d^{3} g^{5} x^{2} - 2 \, d^{3} f g^{4} x - 3 \, d^{3} f^{2} g^{3}\right )} e^{3} - {\left (d^{4} g^{5} x + d^{4} f g^{4}\right )} e^{2}\right )}}, \frac {{\left (c d^{3} g^{3} x + c d^{3} f g^{2} - 3 \, {\left (a g^{3} x^{2} + a f g^{2} x\right )} e^{3} - {\left (4 \, c d f g^{2} x^{2} + 3 \, a d f g^{2} + {\left (4 \, c d f^{2} g + 3 \, a d g^{3}\right )} x\right )} e^{2} + {\left (c d^{2} g^{3} x^{2} - 3 \, c d^{2} f g^{2} x - 4 \, c d^{2} f^{2} g\right )} e\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 (a f^{2} g + {\left (2 \, c f^{3} + 3 \, a f g^{2}\right )} x\right )} e^{4} + {\left (2 \, c d f^{3} + a d f g^{2} - {\left (2 \, c d f^{2} g + 3 \, a d g^{3}\right )} x\right )} e^{3} + {\left (c d^{2} f g^{2} x - c d^{2} f^{2} g - 2 \, a d^{2} g^{3}\right )} e^{2} - {\left (c d^{3} g^{3} x + c d^{3} f g^{2}\right )} e\right )}}{{\left (f^{3} g^{2} x^{2} + f^{4} g x\right )} e^{6} - {\left (3 \, d f^{2} g^{3} x^{2} + 2 \, d f^{3} g^{2} x - d f^{4} g\right )} e^{5} + 3 \, {\left (d^{2} f g^{4} x^{2} - d^{2} f^{3} g^{2}\right )} e^{4} - {\left (d^{3} g^{5} x^{2} - 2 \, d^{3} f g^{4} x - 3 \, d^{3} f^{2} g^{3}\right )} e^{3} - {\left (d^{4} g^{5} x + d^{4} 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)^2/(g*x+f)^(3/2),x, algorithm="fricas")

[Out]

[-1/2*((c*d^3*g^3*x + c*d^3*f*g^2 - 3*(a*g^3*x^2 + a*f*g^2*x)*e^3 - (4*c*d*f*g^2*x^2 + 3*a*d*f*g^2 + (4*c*d*f^
2*g + 3*a*d*g^3)*x)*e^2 + (c*d^2*g^3*x^2 - 3*c*d^2*f*g^2*x - 4*c*d^2*f^2*g)*e)*sqrt(-d*g*e + f*e^2)*log(-(d*g
- (g*x + 2*f)*e - 2*sqrt(-d*g*e + f*e^2)*sqrt(g*x + f))/(x*e + d)) + 2*sqrt(g*x + f)*((a*f^2*g + (2*c*f^3 + 3*
a*f*g^2)*x)*e^4 + (2*c*d*f^3 + a*d*f*g^2 - (2*c*d*f^2*g + 3*a*d*g^3)*x)*e^3 + (c*d^2*f*g^2*x - c*d^2*f^2*g - 2
*a*d^2*g^3)*e^2 - (c*d^3*g^3*x + c*d^3*f*g^2)*e))/((f^3*g^2*x^2 + f^4*g*x)*e^6 - (3*d*f^2*g^3*x^2 + 2*d*f^3*g^
2*x - d*f^4*g)*e^5 + 3*(d^2*f*g^4*x^2 - d^2*f^3*g^2)*e^4 - (d^3*g^5*x^2 - 2*d^3*f*g^4*x - 3*d^3*f^2*g^3)*e^3 -
(d^4*g^5*x + d^4*f*g^4)*e^2), ((c*d^3*g^3*x + c*d^3*f*g^2 - 3*(a*g^3*x^2 + a*f*g^2*x)*e^3 - (4*c*d*f*g^2*x^2
+ 3*a*d*f*g^2 + (4*c*d*f^2*g + 3*a*d*g^3)*x)*e^2 + (c*d^2*g^3*x^2 - 3*c*d^2*f*g^2*x - 4*c*d^2*f^2*g)*e)*sqrt(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)*((a*f^2*g + (2*c*f^3 + 3*
a*f*g^2)*x)*e^4 + (2*c*d*f^3 + a*d*f*g^2 - (2*c*d*f^2*g + 3*a*d*g^3)*x)*e^3 + (c*d^2*f*g^2*x - c*d^2*f^2*g - 2
*a*d^2*g^3)*e^2 - (c*d^3*g^3*x + c*d^3*f*g^2)*e))/((f^3*g^2*x^2 + f^4*g*x)*e^6 - (3*d*f^2*g^3*x^2 + 2*d*f^3*g^
2*x - d*f^4*g)*e^5 + 3*(d^2*f*g^4*x^2 - d^2*f^3*g^2)*e^4 - (d^3*g^5*x^2 - 2*d^3*f*g^4*x - 3*d^3*f^2*g^3)*e^3 -
(d^4*g^5*x + d^4*f*g^4)*e^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + c x^{2}}{\left (d + e x\right )^{2} \left (f + g x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.37, size = 225, normalized size = 1.56 \begin {gather*} \frac {{\left (c d^{2} g - 4 \, c d f e - 3 \, a g e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{{\left (d^{2} g^{2} e - 2 \, d f g e^{2} + f^{2} e^{3}\right )} \sqrt {d g e - f e^{2}}} - \frac {{\left (g x + f\right )} c d^{2} g^{2} + 2 \, c d f^{2} g e + 2 \, a d g^{3} e + 2 \, {\left (g x + f\right )} c f^{2} e^{2} - 2 \, c f^{3} e^{2} + 3 \, {\left (g x + f\right )} a g^{2} e^{2} - 2 \, a f g^{2} e^{2}}{{\left (d^{2} g^{3} e - 2 \, d f g^{2} e^{2} + f^{2} g e^{3}\right )} {\left (\sqrt {g x + f} d g + {\left (g x + f\right )}^{\frac {3}{2}} e - \sqrt {g x + f} f e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*d^2*g - 4*c*d*f*e - 3*a*g*e^2)*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))/((d^2*g^2*e - 2*d*f*g*e^2 + f^2*
e^3)*sqrt(d*g*e - f*e^2)) - ((g*x + f)*c*d^2*g^2 + 2*c*d*f^2*g*e + 2*a*d*g^3*e + 2*(g*x + f)*c*f^2*e^2 - 2*c*f
^3*e^2 + 3*(g*x + f)*a*g^2*e^2 - 2*a*f*g^2*e^2)/((d^2*g^3*e - 2*d*f*g^2*e^2 + f^2*g*e^3)*(sqrt(g*x + f)*d*g +
(g*x + f)^(3/2)*e - sqrt(g*x + f)*f*e))

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Mupad [B]
time = 3.29, size = 187, normalized size = 1.30 \begin {gather*} -\frac {\frac {2\,\left (c\,f^2+a\,g^2\right )}{d\,g-e\,f}+\frac {\left (f+g\,x\right )\,\left (c\,d^2\,g^2+2\,c\,e^2\,f^2+3\,a\,e^2\,g^2\right )}{e\,{\left (d\,g-e\,f\right )}^2}}{\sqrt {f+g\,x}\,\left (d\,g^2-e\,f\,g\right )+e\,g\,{\left (f+g\,x\right )}^{3/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {f+g\,x}\,\left (d^2\,e\,g^2-2\,d\,e^2\,f\,g+e^3\,f^2\right )}{\sqrt {e}\,{\left (d\,g-e\,f\right )}^{5/2}}\right )\,\left (-c\,g\,d^2+4\,c\,f\,d\,e+3\,a\,g\,e^2\right )}{e^{3/2}\,{\left (d\,g-e\,f\right )}^{5/2}} \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)^2),x)

[Out]

- ((2*(a*g^2 + c*f^2))/(d*g - e*f) + ((f + g*x)*(3*a*e^2*g^2 + c*d^2*g^2 + 2*c*e^2*f^2))/(e*(d*g - e*f)^2))/((
f + g*x)^(1/2)*(d*g^2 - e*f*g) + e*g*(f + g*x)^(3/2)) - (atan(((f + g*x)^(1/2)*(e^3*f^2 + d^2*e*g^2 - 2*d*e^2*
f*g))/(e^(1/2)*(d*g - e*f)^(5/2)))*(3*a*e^2*g - c*d^2*g + 4*c*d*e*f))/(e^(3/2)*(d*g - e*f)^(5/2))

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