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

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

[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 = 0.27, 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 = {898, 1259, 453, 208} \[ -\frac {\sqrt {f+g x} \left (a e^2+c d^2\right )}{e (d+e x) (e f-d g)^2}+\frac {\left (3 a e^2 g+c d (4 e f-d g)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{5/2}}-\frac {2 \left (a g^2+c f^2\right )}{g \sqrt {f+g x} (e f-d g)^2} \]

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 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 453

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 898

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 1259

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*(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)))/(d + e*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 \operatorname {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 \operatorname {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 ) \operatorname {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 [C]  time = 0.08, size = 118, normalized size = 0.82 \[ -\frac {2 \left (g^2 \left (a e^2+c d^2\right ) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {e (f+g x)}{e f-d g}\right )+2 c d g (e f-d g) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {e (f+g x)}{e f-d g}\right )+c (e f-d g)^2\right )}{e^2 g \sqrt {f+g x} (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c*(e*f - d*g)^2 + 2*c*d*g*(e*f - d*g)*Hypergeometric2F1[-1/2, 1, 1/2, (e*(f + g*x))/(e*f - d*g)] + (c*d^2
+ a*e^2)*g^2*Hypergeometric2F1[-1/2, 2, 1/2, (e*(f + g*x))/(e*f - d*g)]))/(e^2*g*(e*f - d*g)^2*Sqrt[f + g*x])

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fricas [B]  time = 0.67, size = 906, normalized size = 6.29 \[ \left [\frac {{\left (4 \, c d^{2} e f^{2} g - {\left (c d^{3} - 3 \, a d e^{2}\right )} f g^{2} + {\left (4 \, c d e^{2} f g^{2} - {\left (c d^{2} e - 3 \, a e^{3}\right )} g^{3}\right )} x^{2} + {\left (4 \, c d e^{2} f^{2} g + 3 \, {\left (c d^{2} e + a e^{3}\right )} f g^{2} - {\left (c d^{3} - 3 \, a d e^{2}\right )} g^{3}\right )} x\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 d e^{3} f^{3} - 2 \, a d^{2} e^{2} g^{3} - {\left (c d^{2} e^{2} - a e^{4}\right )} f^{2} g - {\left (c d^{3} e - a d e^{3}\right )} f g^{2} + {\left (2 \, c e^{4} f^{3} - 2 \, c d e^{3} f^{2} g + {\left (c d^{2} e^{2} + 3 \, a e^{4}\right )} f g^{2} - {\left (c d^{3} e + 3 \, a d e^{3}\right )} g^{3}\right )} x\right )} \sqrt {g x + f}}{2 \, {\left (d e^{5} f^{4} g - 3 \, d^{2} e^{4} f^{3} g^{2} + 3 \, d^{3} e^{3} f^{2} g^{3} - d^{4} e^{2} f g^{4} + {\left (e^{6} f^{3} g^{2} - 3 \, d e^{5} f^{2} g^{3} + 3 \, d^{2} e^{4} f g^{4} - d^{3} e^{3} g^{5}\right )} x^{2} + {\left (e^{6} f^{4} g - 2 \, d e^{5} f^{3} g^{2} + 2 \, d^{3} e^{3} f g^{4} - d^{4} e^{2} g^{5}\right )} x\right )}}, -\frac {{\left (4 \, c d^{2} e f^{2} g - {\left (c d^{3} - 3 \, a d e^{2}\right )} f g^{2} + {\left (4 \, c d e^{2} f g^{2} - {\left (c d^{2} e - 3 \, a e^{3}\right )} g^{3}\right )} x^{2} + {\left (4 \, c d e^{2} f^{2} g + 3 \, {\left (c d^{2} e + a e^{3}\right )} f g^{2} - {\left (c d^{3} - 3 \, a d e^{2}\right )} g^{3}\right )} x\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 d e^{3} f^{3} - 2 \, a d^{2} e^{2} g^{3} - {\left (c d^{2} e^{2} - a e^{4}\right )} f^{2} g - {\left (c d^{3} e - a d e^{3}\right )} f g^{2} + {\left (2 \, c e^{4} f^{3} - 2 \, c d e^{3} f^{2} g + {\left (c d^{2} e^{2} + 3 \, a e^{4}\right )} f g^{2} - {\left (c d^{3} e + 3 \, a d e^{3}\right )} g^{3}\right )} x\right )} \sqrt {g x + f}}{d e^{5} f^{4} g - 3 \, d^{2} e^{4} f^{3} g^{2} + 3 \, d^{3} e^{3} f^{2} g^{3} - d^{4} e^{2} f g^{4} + {\left (e^{6} f^{3} g^{2} - 3 \, d e^{5} f^{2} g^{3} + 3 \, d^{2} e^{4} f g^{4} - d^{3} e^{3} g^{5}\right )} x^{2} + {\left (e^{6} f^{4} g - 2 \, d e^{5} f^{3} g^{2} + 2 \, d^{3} e^{3} f g^{4} - d^{4} e^{2} g^{5}\right )} x}\right ] \]

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

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giac [A]  time = 0.19, size = 225, normalized size = 1.56 \[ \frac {{\left (c d^{2} g - 4 \, c d f e - 3 \, a g e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{{\left (d^{2} g^{2} e - 2 \, d f g e^{2} + f^{2} e^{3}\right )} \sqrt {d g e - f e^{2}}} - \frac {{\left (g x + f\right )} c d^{2} g^{2} + 2 \, c d f^{2} g e + 2 \, a d g^{3} e + 2 \, {\left (g x + f\right )} c f^{2} e^{2} - 2 \, c f^{3} e^{2} + 3 \, {\left (g x + f\right )} a g^{2} e^{2} - 2 \, a f g^{2} e^{2}}{{\left (d^{2} g^{3} e - 2 \, d f g^{2} e^{2} + f^{2} g e^{3}\right )} {\left (\sqrt {g x + f} d g + {\left (g x + f\right )}^{\frac {3}{2}} e - \sqrt {g x + f} f e\right )}} \]

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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maple [B]  time = 0.02, size = 269, normalized size = 1.87 \[ -\frac {3 a e g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{2} \sqrt {\left (d g -e f \right ) e}}+\frac {c \,d^{2} g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{2} \sqrt {\left (d g -e f \right ) e}\, e}-\frac {4 c d f \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{2} \sqrt {\left (d g -e f \right ) e}}-\frac {\sqrt {g x +f}\, a e g}{\left (d g -e f \right )^{2} \left (e g x +d g \right )}-\frac {\sqrt {g x +f}\, c \,d^{2} g}{\left (d g -e f \right )^{2} \left (e g x +d g \right ) e}-\frac {2 a g}{\left (d g -e f \right )^{2} \sqrt {g x +f}}-\frac {2 c \,f^{2}}{\left (d g -e f \right )^{2} \sqrt {g x +f}\, g} \]

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)

[Out]

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

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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