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

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

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Rubi [A]  time = 0.10, antiderivative size = 131, 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 = {686, 694, 329, 212, 206, 203} \begin {gather*} -\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} \end {gather*}

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 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 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]

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*}

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Mathematica [A]  time = 0.23, size = 126, normalized size = 0.96 \begin {gather*} 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 ) \end {gather*}

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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IntegrateAlgebraic [C]  time = 0.86, size = 224, normalized size = 1.71 \begin {gather*} \frac {(1-i) c d^{3/2} \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 )}{\left (b^2-4 a c\right )^{3/4}}-\frac {(1-i) c d^{3/2} \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 )}{\left (b^2-4 a c\right )^{3/4}}-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 0.44, size = 549, normalized size = 4.19 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.25, size = 438, normalized size = 3.34 \begin {gather*} \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 {gather*}

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)

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maple [B]  time = 0.06, size = 326, normalized size = 2.49 \begin {gather*} -\frac {\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 {3}{4}}}+\frac {\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 {3}{4}}}+\frac {\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 {3}{4}}}-\frac {4 \sqrt {2 c d x +b d}\, c \,d^{3}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}} \end {gather*}

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 = 0.00, 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((2*c*d*x+b*d)^(3/2)/(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 = 0.62, size = 225, normalized size = 1.72 \begin {gather*} -\frac {4\,c\,d^3\,\sqrt {b\,d+2\,c\,d\,x}}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-\frac {2\,c\,d^{3/2}\,\mathrm {atan}\left (\frac {128\,c^3\,d^{15/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,c^3\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c^4\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {2\,c\,d^{3/2}\,\mathrm {atanh}\left (\frac {128\,c^3\,d^{15/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,c^3\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c^4\,d^8}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (4*c*d^3*(b*d + 2*c*d*x)^(1/2))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) - (2*c*d^(3/2)*atan((128*c^3*d^(15
/2)*(b*d + 2*c*d*x)^(1/2))/(((128*b^2*c^3*d^8)/(b^2 - 4*a*c)^(3/2) - (512*a*c^4*d^8)/(b^2 - 4*a*c)^(3/2))*(b^2
- 4*a*c)^(3/4))))/(b^2 - 4*a*c)^(3/4) - (2*c*d^(3/2)*atanh((128*c^3*d^(15/2)*(b*d + 2*c*d*x)^(1/2))/(((128*b^
2*c^3*d^8)/(b^2 - 4*a*c)^(3/2) - (512*a*c^4*d^8)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(b^2 - 4*a*c)^(3/
4)

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sympy [F(-1)]  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((2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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