3.1303 \(\int \frac{1}{\sqrt{b d+2 c d x} (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=143 \[ -\frac{\sqrt{b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{6 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{7/4}}+\frac{6 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{7/4}} \]

[Out]

-(Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (6*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1
/4)*Sqrt[d])])/((b^2 - 4*a*c)^(7/4)*Sqrt[d]) + (6*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]
)/((b^2 - 4*a*c)^(7/4)*Sqrt[d])

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Rubi [A]  time = 0.102449, antiderivative size = 143, 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 = {687, 694, 329, 212, 206, 203} \[ -\frac{\sqrt{b d+2 c d x}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{6 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{7/4}}+\frac{6 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (6*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1
/4)*Sqrt[d])])/((b^2 - 4*a*c)^(7/4)*Sqrt[d]) + (6*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]
)/((b^2 - 4*a*c)^(7/4)*Sqrt[d])

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && RationalQ[m] && 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{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx &=-\frac{\sqrt{b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{(3 c) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac{\sqrt{b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{3 \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 )}{2 \left (b^2-4 a c\right ) d}\\ &=-\frac{\sqrt{b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{3 \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 )}{\left (b^2-4 a c\right ) d}\\ &=-\frac{\sqrt{b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac{(6 c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d-x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{(6 c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d+x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{3/2}}\\ &=-\frac{\sqrt{b d+2 c d x}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac{6 c \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 )^{7/4} \sqrt{d}}+\frac{6 c \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 )^{7/4} \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.163875, size = 167, normalized size = 1.17 \[ \frac{-\left (b^2-4 a c\right ) (b+2 c x)+6 c \sqrt [4]{b^2-4 a c} \sqrt{b+2 c x} (a+x (b+c x)) \tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+6 c \sqrt [4]{b^2-4 a c} \sqrt{b+2 c x} (a+x (b+c x)) \tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^2 (a+x (b+c x)) \sqrt{d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b^2 - 4*a*c)*(b + 2*c*x)) + 6*c*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x]*(a + x*(b + c*x))*ArcTan[Sqrt[b + 2*c*
x]/(b^2 - 4*a*c)^(1/4)] + 6*c*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x]*(a + x*(b + c*x))*ArcTanh[Sqrt[b + 2*c*x]/(b
^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*(a + x*(b + c*x)))

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Maple [B]  time = 0.191, size = 345, normalized size = 2.4 \begin{align*} 4\,{\frac{c{d}^{3}\sqrt{2\,cdx+bd}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) }}+{\frac{3\,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{7}{4}}}}+3\,{\frac{c{d}^{3}\sqrt{2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{7/4}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }-3\,{\frac{c{d}^{3}\sqrt{2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{7/4}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^2,x)

[Out]

4*c*d^3*(2*c*d*x+b*d)^(1/2)/(4*a*c*d^2-b^2*d^2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)+3/2*c*d^3/(4*a*c*d^2-b^2
*d^2)^(7/4)*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)))+3*c*d^3/
(4*a*c*d^2-b^2*d^2)^(7/4)*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)-3*c*d^3/(4*a
*c*d^2-b^2*d^2)^(7/4)*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(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.22683, size = 2684, normalized size = 18.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12*((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*
b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^
(1/4)*arctan(-((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*sqrt(
2*c^3*d*x + b*c^2*d + (b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*sqrt(c^4/((b^14 - 28
*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 1
6384*a^7*c^7)*d^2))*d^2)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 -
21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(3/4)*d - (b^10*c - 20*a*b^8*c^2 + 160*a^2*b^6*c^
3 - 640*a^3*b^4*c^4 + 1280*a^4*b^2*c^5 - 1024*a^5*c^6)*sqrt(2*c*d*x + b*d)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2
*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))
^(3/4)*d)/c^4) + 3*((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d)*(c^4/((b^14 - 28*a*b^
12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*
a^7*c^7)*d^2))^(1/4)*log(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*
a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(1/4)*d + 3*sqrt
(2*c*d*x + b*d)*c) - 3*((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d)*(c^4/((b^14 - 28*
a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16
384*a^7*c^7)*d^2))^(1/4)*log(-3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c^4/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 -
2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^2))^(1/4)*d + 3
*sqrt(2*c*d*x + b*d)*c) - sqrt(2*c*d*x + b*d))/((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2
*c)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**2,x)

[Out]

Integral(1/(sqrt(d*(b + 2*c*x))*(a + b*x + c*x**2)**2), x)

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Giac [B]  time = 1.16489, size = 684, normalized size = 4.78 \begin{align*} \frac{3 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c \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^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac{3 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c \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^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac{3 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c \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^{4} d - 8 \, \sqrt{2} a b^{2} c d + 16 \, \sqrt{2} a^{2} c^{2} d} - \frac{3 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c \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^{4} d - 8 \, \sqrt{2} a b^{2} c d + 16 \, \sqrt{2} a^{2} c^{2} d} + \frac{4 \, \sqrt{2 \, c d x + b d} c d}{{\left (b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}\right )}{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*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^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d) + 3*sqrt(2)*(-b^2*d^2 + 4*a*c*
d^2)^(1/4)*c*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^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d) + 3*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*log(2*c*d*x + b*d + sq
rt(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^4*d - 8*sqrt(2
)*a*b^2*c*d + 16*sqrt(2)*a^2*c^2*d) - 3*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*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^4*d - 8*sqrt(2)*a*b^2*c*d + 16*
sqrt(2)*a^2*c^2*d) + 4*sqrt(2*c*d*x + b*d)*c*d/((b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)*(b^2 - 4*a*c))