3.12.5 \(\int \frac {1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=131 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, 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 = {693, 694, 329, 212, 206, 203} \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

4/(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) - (2*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((
b^2 - 4*a*c)^(7/4)*d^(5/2)) - (2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(7
/4)*d^(5/2))

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 693

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

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}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx &=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {\int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) d^2}\\ &=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {\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 c \left (b^2-4 a c\right ) d^3}\\ &=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {\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 )}{c \left (b^2-4 a c\right ) d^3}\\ &=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}-\frac {2 \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} d^2}-\frac {2 \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} d^2}\\ &=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}-\frac {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 )^{7/4} d^{5/2}}-\frac {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 )^{7/4} d^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 56, normalized size = 0.43 \begin {gather*} \frac {4 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right ) (d (b+2 c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

(4*Hypergeometric2F1[-3/4, 1, 1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)*d*(d*(b + 2*c*x))^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 0.40, size = 232, normalized size = 1.77 \begin {gather*} \frac {(1-i) \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 )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}-\frac {(1-i) \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 )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}+\frac {4 \sqrt {b d+2 c d x}}{3 d^3 \left (b^2-4 a c\right ) (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

(4*Sqrt[b*d + 2*c*d*x])/(3*(b^2 - 4*a*c)*d^3*(b + 2*c*x)^2) + ((1 - I)*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)*d^(5/2)) - ((1 - I)*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)*d^(5/2))

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 1163, normalized size = 8.88

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

-1/3*(12*(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*a*b^2*c)*d^3)*(1/((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^10))^(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((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)*d^6*sqrt(1/((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^10)) + 2*c*d*x + b*d)*d^7*(1/((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^10))^(3/4) + (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*d*x + b*d)*d^7*(1/((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^10))^(3/4)) + 3*(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*a*b^2*c)*d^3)*
(1/((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^10))^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*(1/((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^
10))^(1/4) + sqrt(2*c*d*x + b*d)) - 3*(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*
a*b^2*c)*d^3)*(1/((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^10))^(1/4)*log(-(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*(1/((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^10))^(1/4) + sqrt(2*c*d*x + b*d)) - 4*sqrt(2*c*d*x + b*d))/(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(
b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*a*b^2*c)*d^3)

________________________________________________________________________________________

giac [B]  time = 0.21, size = 497, normalized size = 3.79 \begin {gather*} -\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \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^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \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^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \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^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \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^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {4}{3 \, {\left (b^{2} d - 4 \, a c d\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) - sqrt(2)*(-b^2*d^2 + 4*a*c
*d^2)^(1/4)*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^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) - (-b^2*d^2 + 4*a*c*d^2)^(1/4)*log(2*c*d*x + b*d + s
qrt(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^3 - 8*sqr
t(2)*a*b^2*c*d^3 + 16*sqrt(2)*a^2*c^2*d^3) + (-b^2*d^2 + 4*a*c*d^2)^(1/4)*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^3 - 8*sqrt(2)*a*b^2*c*d^
3 + 16*sqrt(2)*a^2*c^2*d^3) + 4/3/((b^2*d - 4*a*c*d)*(2*c*d*x + b*d)^(3/2))

________________________________________________________________________________________

maple [B]  time = 0.06, size = 341, normalized size = 2.60 \begin {gather*} \frac {\sqrt {2}\, \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 -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}} d}-\frac {\sqrt {2}\, \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 -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}} d}-\frac {\sqrt {2}\, \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 -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}} d}-\frac {4}{3 \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x)

[Out]

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

________________________________________________________________________________________

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)^(5/2)/(c*x^2+b*x+a),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?

________________________________________________________________________________________

mupad [B]  time = 0.70, size = 1345, normalized size = 10.27

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x)

[Out]

4/(3*(b*d + 2*c*d*x)^(3/2)*(b^2*d - 4*a*c*d)) + (atan(((((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^3 - 4096*a^3*c^3*d^3
+ 3072*a^2*b^2*c^2*d^3 - 768*a*b^4*c*d^3) + (64*b^10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2*b^6*c^2*d^6 - 40960*a
^3*b^4*c^3*d^6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^8*c*d^6)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))*1i)/(d^(5/2)*(b^2 -
4*a*c)^(7/4)) + (((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^3 - 4096*a^3*c^3*d^3 + 3072*a^2*b^2*c^2*d^3 - 768*a*b^4*c*d^
3) - (64*b^10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*d^6 + 81920*a^4*b^2*c^4*d^6
- 1280*a*b^8*c*d^6)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))*1i)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))/(((b*d + 2*c*d*x)^(1/2)*
(64*b^6*d^3 - 4096*a^3*c^3*d^3 + 3072*a^2*b^2*c^2*d^3 - 768*a*b^4*c*d^3) + (64*b^10*d^6 - 65536*a^5*c^5*d^6 +
10240*a^2*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*d^6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^8*c*d^6)/(d^(5/2)*(b^2 - 4*a*
c)^(7/4)))/(d^(5/2)*(b^2 - 4*a*c)^(7/4)) - ((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^3 - 4096*a^3*c^3*d^3 + 3072*a^2*b^
2*c^2*d^3 - 768*a*b^4*c*d^3) - (64*b^10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*d^
6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^8*c*d^6)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))/(d^(5/2)*(b^2 - 4*a*c)^(7/4))))*2
i)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)) - (2*atan((((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^3 - 4096*a^3*c^3*d^3 + 3072*a^2*b
^2*c^2*d^3 - 768*a*b^4*c*d^3) - ((64*b^10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*
d^6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^8*c*d^6)*1i)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))/(d^(5/2)*(b^2 - 4*a*c)^(7/4
)) + ((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^3 - 4096*a^3*c^3*d^3 + 3072*a^2*b^2*c^2*d^3 - 768*a*b^4*c*d^3) + ((64*b^
10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*d^6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^
8*c*d^6)*1i)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))/((((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^
3 - 4096*a^3*c^3*d^3 + 3072*a^2*b^2*c^2*d^3 - 768*a*b^4*c*d^3) - ((64*b^10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2
*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*d^6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^8*c*d^6)*1i)/(d^(5/2)*(b^2 - 4*a*c)^(7
/4)))*1i)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)) - (((b*d + 2*c*d*x)^(1/2)*(64*b^6*d^3 - 4096*a^3*c^3*d^3 + 3072*a^2*b^
2*c^2*d^3 - 768*a*b^4*c*d^3) + ((64*b^10*d^6 - 65536*a^5*c^5*d^6 + 10240*a^2*b^6*c^2*d^6 - 40960*a^3*b^4*c^3*d
^6 + 81920*a^4*b^2*c^4*d^6 - 1280*a*b^8*c*d^6)*1i)/(d^(5/2)*(b^2 - 4*a*c)^(7/4)))*1i)/(d^(5/2)*(b^2 - 4*a*c)^(
7/4)))))/(d^(5/2)*(b^2 - 4*a*c)^(7/4))

________________________________________________________________________________________

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(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________