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

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

[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])

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Rubi [A]  time = 1.35614, antiderivative size = 429, normalized size of antiderivative = 1., 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 = {911, 96, 93, 208} \[ -\frac{8 c^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{f+g x} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )^{3/2} \sqrt{2 c f-g \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{8 c^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{f+g x} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )^{3/2} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{4 c e \sqrt{f+g x}}{\sqrt{b^2-4 a c} \sqrt{d+e x} (e f-d g) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{4 c e \sqrt{f+g x}}{\sqrt{b^2-4 a c} \sqrt{d+e x} (e f-d g) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )} \]

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 911

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]

Rule 96

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 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 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]

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 ) \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 )}{\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 ) \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 )}{\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*}

Mathematica [A]  time = 2.11857, size = 340, normalized size = 0.79 \[ \frac{4 c \left (\frac{e^2 \sqrt{b^2-4 a c} \sqrt{f+g x}}{2 c \sqrt{d+e x} (d g-e f) \left (e (a e-b d)+c d^2\right )}-\frac{2 c \tan ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{g \sqrt{b^2-4 a c}-b g+2 c f}}{\sqrt{f+g x} \sqrt{-e \sqrt{b^2-4 a c}+b e-2 c d}}\right )}{\left (e \left (b-\sqrt{b^2-4 a c}\right )-2 c d\right )^{3/2} \sqrt{g \left (\sqrt{b^2-4 a c}-b\right )+2 c f}}+\frac{2 c \tan ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{f+g x} \sqrt{e \left (\sqrt{b^2-4 a c}+b\right )-2 c d}}\right )}{\left (e \left (\sqrt{b^2-4 a c}+b\right )-2 c d\right )^{3/2} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.916, size = 47351, normalized size = 110.4 \begin{align*} \text{output too large to display} \end{align*}

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)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}{\left (e x + d\right )}^{\frac{3}{2}} \sqrt{g x + f}}\,{d x} \end{align*}

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)*(e*x + d)^(3/2)*sqrt(g*x + f)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError