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

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

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Rubi [A]  time = 0.10, antiderivative size = 143, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.16, size = 167, normalized size = 1.17 \begin {gather*} \frac {-\left (\left (b^2-4 a c\right ) (b+2 c x)\right )+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)}} \end {gather*}

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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IntegrateAlgebraic [C]  time = 0.41, size = 236, normalized size = 1.65 \begin {gather*} -\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 {(3-3 i) c \tan ^{-1}\left (\frac {-\frac {(1+i) c \sqrt {d} x}{\sqrt [4]{b^2-4 a c}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {d}}{\sqrt [4]{b^2-4 a c}}+\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {d} \sqrt [4]{b^2-4 a c}}{\sqrt {b d+2 c d x}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{7/4}}+\frac {(3-3 i) c \tanh ^{-1}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b d+2 c d x}}{\sqrt {d} \left (\sqrt {b^2-4 a c}+i b+2 i c x\right )}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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))) - ((3 - 3*I)*c*ArcTan[(((-1/2 - I/2)*b*Sqrt[d])/(b^
2 - 4*a*c)^(1/4) + (1/2 - I/2)*(b^2 - 4*a*c)^(1/4)*Sqrt[d] - ((1 + I)*c*Sqrt[d]*x)/(b^2 - 4*a*c)^(1/4))/Sqrt[b
*d + 2*c*d*x]])/((b^2 - 4*a*c)^(7/4)*Sqrt[d]) + ((3 - 3*I)*c*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b*d + 2
*c*d*x])/(Sqrt[d]*(I*b + Sqrt[b^2 - 4*a*c] + (2*I)*c*x))])/((b^2 - 4*a*c)^(7/4)*Sqrt[d])

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fricas [B]  time = 0.46, size = 1147, normalized size = 8.02

result too large to display

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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giac [B]  time = 0.22, size = 507, normalized size = 3.55 \begin {gather*} \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 {gather*}

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))

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maple [B]  time = 0.06, size = 345, normalized size = 2.41 \begin {gather*} -\frac {3 \sqrt {2}\, c \,d^{3} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {7}{4}}}+\frac {3 \sqrt {2}\, c \,d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {7}{4}}}+\frac {3 \sqrt {2}\, c \,d^{3} \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {7}{4}}}+\frac {4 \sqrt {2 c d x +b d}\, c \,d^{3}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 0.58, size = 183, normalized size = 1.28 \begin {gather*} \frac {6\,c\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{7/4}}+\frac {6\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{7/4}}+\frac {4\,c\,d\,\sqrt {b\,d+2\,c\,d\,x}}{\left (4\,a\,c-b^2\right )\,\left ({\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*c*atan(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4))/(d^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))))/(d^(1/2)*(b^2
- 4*a*c)^(7/4)) + (6*c*atanh(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(7/4))/(d^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2
*c))))/(d^(1/2)*(b^2 - 4*a*c)^(7/4)) + (4*c*d*(b*d + 2*c*d*x)^(1/2))/((4*a*c - b^2)*((b*d + 2*c*d*x)^2 - b^2*d
^2 + 4*a*c*d^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {d \left (b + 2 c x\right )} \left (a + b x + c x^{2}\right )^{2}}\, dx \end {gather*}

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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