\(\int \frac {1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)} \, dx\) [1293]
Optimal result
Integrand size = 26, antiderivative size = 131 \[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}-\frac {2 \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} d^{5/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} d^{5/2}}
\]
[Out]
4/3/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(3/2)-2*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(
7/4)/d^(5/2)-2*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(7/4)/d^(5/2)
Rubi [A] (verified)
Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {707, 708, 335, 218, 212, 209}
\[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}
\]
[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 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 218
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] && !Gt
Q[a/b, 0]
Rule 335
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 707
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 708
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*}
\text {integral}& = \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 {\text {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 {\text {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 \text {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 \text {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] (verified)
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.85
\[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\frac {\left (\frac {1}{3}-\frac {i}{3}\right ) \left ((2+2 i) \left (b^2-4 a c\right )^{3/4}+3 (b+2 c x)^{3/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-3 b \sqrt {b+2 c x} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-6 c x \sqrt {b+2 c x} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-3 (b+2 c x)^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{7/4} d (d (b+2 c x))^{3/2}}
\]
[In]
Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x]
[Out]
((1/3 - I/3)*((2 + 2*I)*(b^2 - 4*a*c)^(3/4) + 3*(b + 2*c*x)^(3/2)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 -
4*a*c)^(1/4)] - 3*b*Sqrt[b + 2*c*x]*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)] - 6*c*x*Sqrt[b +
2*c*x]*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)] - 3*(b + 2*c*x)^(3/2)*ArcTanh[((1 + I)*(b^2
- 4*a*c)^(1/4)*Sqrt[b + 2*c*x])/(Sqrt[b^2 - 4*a*c] + I*(b + 2*c*x))]))/((b^2 - 4*a*c)^(7/4)*d*(d*(b + 2*c*x))^
(3/2))
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(273\) vs. \(2(107)=214\).
Time = 2.50 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.09
| | |
| method | result | size |
| | |
| derivativedivides |
\(4 d \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{2} \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {1}{3 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}\right )\) |
\(274\) |
| default |
\(4 d \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{2} \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {1}{3 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}\right )\) |
\(274\) |
| pseudoelliptic |
\(-\frac {d \sqrt {2}\, \left (2 c x +b \right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right ) \sqrt {d \left (2 c x +b \right )}+\frac {8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}{3}}{8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \sqrt {d \left (2 c x +b \right )}\, d^{2} \left (2 c x +b \right ) \left (-\frac {b^{2}}{4}+a c \right )}\) |
\(310\) |
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[In]
int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
[Out]
4*d*(-1/8/d^2/(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)))+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)-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))-1/3/d^2/(4*a*c-b^2)/(2*c*d*x+b*d)^(3/2))
Fricas [C] (verification not implemented)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 1116, normalized size of antiderivative = 8.52
\[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display}
\]
[In]
integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")
[Out]
-1/3*(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 - 163
84*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 - 2
240*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) + sqr
t(2*c*d*x + b*d)) - 3*(-4*I*(b^2*c^2 - 4*a*c^3)*d^3*x^2 - 4*I*(b^3*c - 4*a*b*c^2)*d^3*x - I*(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 + 28
672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^10))^(1/4)*log(I*(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*I*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*I*(b^3*c - 4*a*b*c^2)*d^3*x
+ I*(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(-I*(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 + 89
60*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 + 1
6*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)) - 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)
Sympy [F(-1)]
Timed out. \[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Timed out}
\]
[In]
integrate(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a),x)
[Out]
Timed out
Maxima [F(-2)]
Exception generated. \[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError}
\]
[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 deta
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 497 vs. \(2 (107) = 214\).
Time = 0.29 (sec) , antiderivative size = 497, normalized size of antiderivative = 3.79
\[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=-\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {4}{3 \, {\left (b^{2} d - 4 \, a c d\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}
\]
[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))
Mupad [B] (verification not implemented)
Time = 9.53 (sec) , antiderivative size = 1345, normalized size of antiderivative = 10.27
\[
\int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display}
\]
[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))