3.9.53 \(\int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} (a+b x+c x^2)} \, dx\) [853]

Optimal. Leaf size=429 \[ \frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {8 c^2 \tanh ^{-1}\left (\frac {\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right )^{3/2} \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}+\frac {8 c^2 \tanh ^{-1}\left (\frac {\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right )^{3/2} \sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \]

[Out]

4*c*e*(g*x+f)^(1/2)/(-d*g+e*f)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^2)^(1/2)/(e*x+d)^(1/2)-4*c*e*(g*x+f)
^(1/2)/(-d*g+e*f)/(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))/(e*x+d)^(1/2)-8*c^2*arctanh((e*x+d)^(1/2
)*(2*c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(g*x+f)^(1/2)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1/2))/(2*c*d-e*(b-(-4
*a*c+b^2)^(1/2)))^(3/2)/(-4*a*c+b^2)^(1/2)/(2*c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2)+8*c^2*arctanh((e*x+d)^(1/2)*
(2*c*f-g*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(g*x+f)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2))/(-4*a*c+b^2)^(1/2
)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(3/2)/(2*c*f-g*(b+(-4*a*c+b^2)^(1/2)))^(1/2)

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Rubi [A]
time = 0.91, antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {925, 98, 95, 214} \begin {gather*} -\frac {8 c^2 \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 )}{\sqrt {b^2-4 a c} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )^{3/2} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}+\frac {8 c^2 \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 )}{\sqrt {b^2-4 a c} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )^{3/2} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}+\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \sqrt {d+e x} (e f-d g) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \sqrt {d+e x} (e f-d g) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*Sqrt[f + g*x]*(a + b*x + c*x^2)),x]

[Out]

(4*c*e*Sqrt[f + g*x])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(e*f - d*g)*Sqrt[d + e*x]) - (4*c
*e*Sqrt[f + g*x])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(e*f - d*g)*Sqrt[d + e*x]) - (8*c^2*A
rcTanh[(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])])/(Sqrt[b^2 - 4*a*c]*(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
]) + (8*c^2*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])])/(Sqrt[b^2 - 4*a*c]*(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])

Rule 95

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 925

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

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac {2 c}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}+2 c x\right ) (d+e x)^{3/2} \sqrt {f+g x}}-\frac {2 c}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}+2 c x\right ) (d+e x)^{3/2} \sqrt {f+g x}}\right ) \, dx\\ &=\frac {(2 c) \int \frac {1}{\left (b-\sqrt {b^2-4 a c}+2 c x\right ) (d+e x)^{3/2} \sqrt {f+g x}} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {1}{\left (b+\sqrt {b^2-4 a c}+2 c x\right ) (d+e x)^{3/2} \sqrt {f+g x}} \, dx}{\sqrt {b^2-4 a c}}\\ &=\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}+\frac {\left (4 c^2\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}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right )}-\frac {\left (4 c^2\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}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right )}\\ &=\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}+\frac {\left (8 c^2\right ) \text {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 )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right )}-\frac {\left (8 c^2\right ) \text {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 )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right )}\\ &=\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {8 c^2 \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 )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right )^{3/2} \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}+\frac {8 c^2 \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 )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right )^{3/2} \sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}}\\ \end {align*}

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Mathematica [A]
time = 3.75, size = 543, normalized size = 1.27 \begin {gather*} \frac {2 e^2 \sqrt {f+g x}}{\left (c d^2+e (-b d+a e)\right ) (-e f+d g) \sqrt {d+e x}}+\frac {\sqrt {2} \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right ) \tan ^{-1}\left (\frac {\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}}\right )}{\sqrt {b^2-4 a c} \left (c d^2+e (-b d+a e)\right )^{3/2} \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}}+\frac {\sqrt {2} \left (-2 c^2 d^2+b \left (-b+\sqrt {b^2-4 a c}\right ) e^2+2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right ) \tan ^{-1}\left (\frac {\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}}\right )}{\sqrt {b^2-4 a c} \left (c d^2+e (-b d+a e)\right )^{3/2} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*Sqrt[f + g*x]*(a + b*x + c*x^2)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(47350\) vs. \(2(369)=738\).
time = 0.23, size = 47351, normalized size = 110.38

method	result	size
default	\(\text {Expression too large to display}\)	\(47351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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(1/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {f + g x} \left (a + b x + c x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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