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

Optimal. Leaf size=417 \[ -\frac {2 \left (\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}+e (2 c d-b e)\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{c \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}-\frac {2 \left (e (2 c d-b e)-\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{c \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}+\frac {2 e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{c \sqrt {g}} \]

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Rubi [A]  time = 3.14, antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {909, 63, 217, 206, 6728, 93, 208} \begin {gather*} -\frac {2 \left (\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}+e (2 c d-b e)\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{c \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}-\frac {2 \left (e (2 c d-b e)-\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{c \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}+\frac {2 e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{c \sqrt {g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 909

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Int
[ExpandIntegrand[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b
, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 1.79, size = 401, normalized size = 0.96 \begin {gather*} \frac {\left (e \left (b-\sqrt {b^2-4 a c}\right )-2 c d\right )^{3/2} \sqrt {g \left (\sqrt {b^2-4 a c}+b\right )-2 c f} \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {g \left (b-\sqrt {b^2-4 a c}\right )-2 c f}}{\sqrt {f+g x} \sqrt {e \left (b-\sqrt {b^2-4 a c}\right )-2 c d}}\right )-\left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right )^{3/2} \sqrt {g \left (b-\sqrt {b^2-4 a c}\right )-2 c f} \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}}{\sqrt {f+g x} \sqrt {e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}}\right )}{c \sqrt {b^2-4 a c} \sqrt {g \left (b-\sqrt {b^2-4 a c}\right )-2 c f} \sqrt {g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}}+\frac {2 (e f-d g)^{3/2} \left (\frac {e (f+g x)}{e f-d g}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e f-d g}}\right )}{c \sqrt {g} (f+g x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*f - d*g)^(3/2)*((e*(f + g*x))/(e*f - d*g))^(3/2)*ArcSinh[(Sqrt[g]*Sqrt[d + e*x])/Sqrt[e*f - d*g]])/(c*Sq
rt[g]*(f + g*x)^(3/2)) + ((-2*c*d + (b - Sqrt[b^2 - 4*a*c])*e)^(3/2)*Sqrt[-2*c*f + (b + Sqrt[b^2 - 4*a*c])*g]*
ArcTanh[(Sqrt[-2*c*f + (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(Sqrt[-2*c*d + (b - Sqrt[b^2 - 4*a*c])*e]*Sqr
t[f + g*x])] - (-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)^(3/2)*Sqrt[-2*c*f + (b - Sqrt[b^2 - 4*a*c])*g]*ArcTanh[(Sq
rt[-2*c*f + (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(Sqrt[-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])
])/(c*Sqrt[b^2 - 4*a*c]*Sqrt[-2*c*f + (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[-2*c*f + (b + Sqrt[b^2 - 4*a*c])*g])

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IntegrateAlgebraic [A]  time = 6.48, size = 479, normalized size = 1.15 \begin {gather*} \frac {\sqrt {2} \left (e \sqrt {b^2-4 a c}-b e+2 c d\right ) \sqrt {a e^2-b d e+c d^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {f+g x} \sqrt {a e^2-b d e+c d^2}}{\sqrt {d+e x} \sqrt {-d g \sqrt {b^2-4 a c}+e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}\right )}{c \sqrt {b^2-4 a c} \sqrt {-d g \sqrt {b^2-4 a c}+e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}+\frac {\sqrt {2} \left (e \sqrt {b^2-4 a c}+b e-2 c d\right ) \sqrt {a e^2-b d e+c d^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {f+g x} \sqrt {a e^2-b d e+c d^2}}{\sqrt {d+e x} \sqrt {d g \sqrt {b^2-4 a c}-e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}\right )}{c \sqrt {b^2-4 a c} \sqrt {d g \sqrt {b^2-4 a c}-e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}+\frac {2 e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{c \sqrt {g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.18, size = 11688, normalized size = 28.03 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^{3/2}}{\sqrt {f+g\,x}\,\left (c\,x^2+b\,x+a\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)/(c*x**2+b*x+a)/(g*x+f)**(1/2),x)

[Out]

Timed out

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