\(\int \frac {1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)} \, dx\) [117]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\frac {4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}-\frac {2 \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} d^{5/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{7/4} d^{5/2}} \] Output:

4/3/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(3/2)-2*arctan((2*c*d*x+b*d)^(1/2)/(-4*a* 
c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(7/4)/d^(5/2)-2*arctanh((2*c*d*x+b*d)^( 
1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(7/4)/d^(5/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.82 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\frac {\left (\frac {1}{3}-\frac {i}{3}\right ) \left ((2+2 i) \left (b^2-4 a c\right )^{3/4}+3 (b+2 c x)^{3/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-3 b \sqrt {b+2 c x} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-6 c x \sqrt {b+2 c x} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-3 (b+2 c x)^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{7/4} d (d (b+2 c x))^{3/2}} \] Input:

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

Output:

((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]*Ar 
cTan[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 
))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1117, 1118, 27, 25, 266, 756, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1117

\(\displaystyle \frac {\int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 1118

\(\displaystyle \frac {\int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)}{2 c d^3 \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}+\frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {2 \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{d \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{d \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac {4 \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )}{d \left (b^2-4 a c\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 1117

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> 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] + Simp[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] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & 
& (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) 
/2])
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[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] && EqQ[2*c*d - b*e, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(273\) vs. \(2(109)=218\).

Time = 1.46 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.06

method	result	size
derivativedivides	\(4 d \left (-\frac {1}{3 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{2} \left (4 a c -b^{2}\right ) \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(274\)
default	\(4 d \left (-\frac {1}{3 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{2} \left (4 a c -b^{2}\right ) \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(274\)
pseudoelliptic	\(-\frac {\frac {3 d \sqrt {2}\, \left (2 c x +b \right ) \left (\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right ) \sqrt {d \left (2 c x +b \right )}}{8}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}{3 \sqrt {d \left (2 c x +b \right )}\, \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (2 c x +b \right ) \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\)	\(310\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 1116, normalized size of antiderivative = 8.39 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-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 - 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 - 215 
04*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^1 
0*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 + 896 
0*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^1 
0))^(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 - 224 
0*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/...
 

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}} \left (a + b x + c x^{2}\right )}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 497 vs. \(2 (109) = 218\).

Time = 0.15 (sec) , antiderivative size = 497, normalized size of antiderivative = 3.74 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=-\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{4} d^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {4}{3 \, {\left (b^{2} d - 4 \, a c d\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}} \] Input:

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

Output:

-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*sq 
rt(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 + 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) 
 + (-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 = 5.71 (sec) , antiderivative size = 1345, normalized size of antiderivative = 10.11 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

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 - 40 
96*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))))*2i)/(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)...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 604, normalized size of antiderivative = 4.54 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx=\frac {\sqrt {d}\, \left (6 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}-2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right ) b +12 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}-2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right ) c x -6 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right ) b -12 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right ) c x +3 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right ) b +6 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right ) c x -3 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right ) b -6 \sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right ) c x -32 a c +8 b^{2}\right )}{6 \sqrt {2 c x +b}\, d^{3} \left (32 a^{2} c^{3} x -16 a \,b^{2} c^{2} x +2 b^{4} c x +16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right )} \] Input:

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

Output:

(sqrt(d)*(6*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b 
**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))* 
b + 12*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)* 
*(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*c*x - 
 6*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/ 
4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*b - 12*sq 
rt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sq 
rt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*c*x + 3*sqrt(b 
 + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4*a*c - b* 
*2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*b + 6*sqrt(b + 2*c*x) 
*(4*a*c - b**2)**(1/4)*sqrt(2)*log( - sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4 
)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*c*x - 3*sqrt(b + 2*c*x)*(4*a*c 
 - b**2)**(1/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) 
+ sqrt(4*a*c - b**2) + b + 2*c*x)*b - 6*sqrt(b + 2*c*x)*(4*a*c - b**2)**(1 
/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c 
 - b**2) + b + 2*c*x)*c*x - 32*a*c + 8*b**2))/(6*sqrt(b + 2*c*x)*d**3*(16* 
a**2*b*c**2 + 32*a**2*c**3*x - 8*a*b**3*c - 16*a*b**2*c**2*x + b**5 + 2*b* 
*4*c*x))