3.1301 \(\int \frac{(b d+2 c d x)^{3/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=131 \[ -\frac{2 c d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac{2 c d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac{d \sqrt{b d+2 c d x}}{a+b x+c x^2} \]

[Out]

-((d*Sqrt[b*d + 2*c*d*x])/(a + b*x + c*x^2)) - (2*c*d^(3/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sq
rt[d])])/(b^2 - 4*a*c)^(3/4) - (2*c*d^(3/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 -
4*a*c)^(3/4)

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Rubi [A]  time = 0.0993918, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {686, 694, 329, 212, 206, 203} \[ -\frac{2 c d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac{2 c d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac{d \sqrt{b d+2 c d x}}{a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Sqrt[b*d + 2*c*d*x])/(a + b*x + c*x^2)) - (2*c*d^(3/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sq
rt[d])])/(b^2 - 4*a*c)^(3/4) - (2*c*d^(3/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 -
4*a*c)^(3/4)

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2
*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.23111, size = 126, normalized size = 0.96 \[ d \sqrt{d (b+2 c x)} \left (-\frac{2 c \tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt{b+2 c x}}-\frac{2 c \tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt{b+2 c x}}-\frac{1}{a+x (b+c x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

d*Sqrt[d*(b + 2*c*x)]*(-(a + x*(b + c*x))^(-1) - (2*c*ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a
*c)^(3/4)*Sqrt[b + 2*c*x]) - (2*c*ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^(3/4)*Sqrt[b +
2*c*x]))

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Maple [B]  time = 0.198, size = 326, normalized size = 2.5 \begin{align*} -4\,{\frac{c{d}^{3}\sqrt{2\,cdx+bd}}{4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2}}}+{\frac{c{d}^{3}\sqrt{2}}{2}\ln \left ({ \left ( 2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) \left ( 2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) ^{-1}} \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}+{c{d}^{3}\sqrt{2}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-{c{d}^{3}\sqrt{2}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x)

[Out]

-4*c*d^3*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)+1/2*c*d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)
*ln((2*c*d*x+b*d+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d
-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2)))+c*d^3/(4*a*c*d^2-b^2*d^2)^(
3/4)*2^(1/2)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-c*d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2
^(1/2)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.97755, size = 1197, normalized size = 9.14 \begin{align*} \frac{4 \, \left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \arctan \left (-\frac{\left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{3}{4}}{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} \sqrt{2 \, c d x + b d} d - \left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{3}{4}} \sqrt{2 \, c^{3} d^{3} x + b c^{2} d^{3} + \sqrt{\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}}{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{c^{4} d^{6}}\right ) - \left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \log \left (\sqrt{2 \, c d x + b d} c d + \left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{1}{4}}{\left (b^{2} - 4 \, a c\right )}\right ) + \left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \log \left (\sqrt{2 \, c d x + b d} c d - \left (\frac{c^{4} d^{6}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac{1}{4}}{\left (b^{2} - 4 \, a c\right )}\right ) - \sqrt{2 \, c d x + b d} d}{c x^{2} + b x + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c*x^2 + b*x + a)*arctan(-((c^4*d^6/(b^6 -
12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(3/4)*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d - (
c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(3/4)*sqrt(2*c^3*d^3*x + b*c^2*d^3 + sqrt(c^4*d^6/(b
^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))*(b^4 - 8*a*b^2*c + 16*a^2*c^2))*(b^4 - 8*a*b^2*c + 16*a^2*c^2)
)/(c^4*d^6)) - (c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c*x^2 + b*x + a)*log(sqrt(2*c
*d*x + b*d)*c*d + (c^4*d^6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(b^2 - 4*a*c)) + (c^4*d^6/(
b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c*x^2 + b*x + a)*log(sqrt(2*c*d*x + b*d)*c*d - (c^4*d^
6/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(b^2 - 4*a*c)) - sqrt(2*c*d*x + b*d)*d)/(c*x^2 + b*x
+ a)

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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((2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.17054, size = 591, normalized size = 4.51 \begin{align*} \frac{4 \, \sqrt{2 \, c d x + b d} c d^{3}}{b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}} - \frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} + 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} - 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d \log \left (2 \, c d x + b d + \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt{2} b^{2} - 4 \, \sqrt{2} a c} + \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d \log \left (2 \, c d x + b d - \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt{2} b^{2} - 4 \, \sqrt{2} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(2*c*d*x + b*d)*c*d^3/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2) - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c
*d*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1
/4))/(b^2 - 4*a*c) - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d
^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^2 - 4*a*c) - (-b^2*d^2 + 4*a*c*d^2)^(1/4)*
c*d*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))
/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + (-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d*log(2*c*d*x + b*d - sqrt(2)*(-b^2*d^2 + 4*a*
c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c)