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

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

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Rubi [A]  time = 0.22, 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 = {898, 1261, 205} \begin {gather*} -\frac {2 \left (a e^2+c d^2\right )}{e^2 \sqrt {d+e x} (e f-d g)}-\frac {2 \left (a g^2+c f^2\right ) \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e f-d g}}\right )}{g^{3/2} (e f-d g)^{3/2}}+\frac {2 c \sqrt {d+e x}}{e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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

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Mathematica [C]  time = 0.09, size = 91, normalized size = 0.81 \begin {gather*} \frac {2 c (e f-d g) (2 d g+e (f+g x))-2 e^2 \left (a g^2+c f^2\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {g (d+e x)}{d g-e f}\right )}{e^2 g^2 \sqrt {d+e x} (e f-d g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.22, size = 118, normalized size = 1.05 \begin {gather*} -\frac {2 \left (a e^2 g+c d^2 g-c e f (d+e x)+c d g (d+e x)\right )}{e^2 g \sqrt {d+e x} (e f-d g)}-\frac {2 \left (a g^2+c f^2\right ) \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e f-d g}}\right )}{g^{3/2} (e f-d g)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 499, normalized size = 4.46 \begin {gather*} \left [\frac {{\left (c d e^{2} f^{2} + a d e^{2} g^{2} + {\left (c e^{3} f^{2} + a e^{3} g^{2}\right )} x\right )} \sqrt {-e f g + d g^{2}} \log \left (\frac {e g x - e f + 2 \, d g - 2 \, \sqrt {-e f g + d g^{2}} \sqrt {e x + d}}{g x + f}\right ) + 2 \, {\left (c d e^{2} f^{2} g - {\left (3 \, c d^{2} e + a e^{3}\right )} f g^{2} + {\left (2 \, c d^{3} + a d e^{2}\right )} g^{3} + {\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 {e x + d}}{d e^{4} f^{2} g^{2} - 2 \, d^{2} e^{3} f g^{3} + d^{3} e^{2} g^{4} + {\left (e^{5} f^{2} g^{2} - 2 \, d e^{4} f g^{3} + d^{2} e^{3} g^{4}\right )} x}, \frac {2 \, {\left ({\left (c d e^{2} f^{2} + a d e^{2} g^{2} + {\left (c e^{3} f^{2} + a e^{3} g^{2}\right )} x\right )} \sqrt {e f g - d g^{2}} \arctan \left (\frac {\sqrt {e f g - d g^{2}} \sqrt {e x + d}}{e g x + d g}\right ) + {\left (c d e^{2} f^{2} g - {\left (3 \, c d^{2} e + a e^{3}\right )} f g^{2} + {\left (2 \, c d^{3} + a d e^{2}\right )} g^{3} + {\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 {e x + d}\right )}}{d e^{4} f^{2} g^{2} - 2 \, d^{2} e^{3} f g^{3} + d^{3} e^{2} g^{4} + {\left (e^{5} f^{2} g^{2} - 2 \, d e^{4} f g^{3} + d^{2} e^{3} g^{4}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 116, normalized size = 1.04 \begin {gather*} \frac {2 \, \sqrt {x e + d} c e^{\left (-2\right )}}{g} + \frac {2 \, {\left (c f^{2} + a g^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} g}{\sqrt {-d g^{2} + f g e}}\right )}{{\left (d g^{2} - f g e\right )} \sqrt {-d g^{2} + f g e}} + \frac {2 \, {\left (c d^{2} + a e^{2}\right )}}{{\left (d g e^{2} - f e^{3}\right )} \sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 114, normalized size = 1.02 \begin {gather*} \frac {-\frac {2 \left (a \,g^{2}+c \,f^{2}\right ) e^{2} \arctanh \left (\frac {\sqrt {e x +d}\, g}{\sqrt {\left (d g -e f \right ) g}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) g}\, g}+\frac {2 \sqrt {e x +d}\, c}{g}-\frac {2 \left (-a \,e^{2}-c \,d^{2}\right )}{\left (d g -e f \right ) \sqrt {e x +d}}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e^2*(c/g*(e*x+d)^(1/2)-e^2*(a*g^2+c*f^2)/(d*g-e*f)/g/((d*g-e*f)*g)^(1/2)*arctanh(g*(e*x+d)^(1/2)/((d*g-e*f)*
g)^(1/2))-(-a*e^2-c*d^2)/(d*g-e*f)/(e*x+d)^(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)^(3/2)/(g*x+f),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(g*(d*g-e*f)>0)', see `assume?`
for more details)Is g*(d*g-e*f)    positive or negative?

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mupad [B]  time = 0.23, size = 124, normalized size = 1.11 \begin {gather*} \frac {2\,c\,\sqrt {d+e\,x}}{e^2\,g}+\frac {2\,\left (c\,g\,d^2+a\,g\,e^2\right )}{e^2\,g\,\left (d\,g-e\,f\right )\,\sqrt {d+e\,x}}+\frac {\mathrm {atan}\left (\frac {d\,g^{3/2}\,\sqrt {d+e\,x}\,1{}\mathrm {i}-e\,f\,\sqrt {g}\,\sqrt {d+e\,x}\,1{}\mathrm {i}}{{\left (d\,g-e\,f\right )}^{3/2}}\right )\,\left (c\,f^2+a\,g^2\right )\,2{}\mathrm {i}}{g^{3/2}\,{\left (d\,g-e\,f\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 87.88, size = 107, normalized size = 0.96 \begin {gather*} \frac {2 c \sqrt {d + e x}}{e^{2} g} + \frac {2 \left (a g^{2} + c f^{2}\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- \frac {d g - e f}{g}}} \right )}}{g^{2} \sqrt {- \frac {d g - e f}{g}} \left (d g - e f\right )} + \frac {2 \left (a e^{2} + c d^{2}\right )}{e^{2} \sqrt {d + e x} \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)**(3/2)/(g*x+f),x)

[Out]

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

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