\(\int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\) [112]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {4}{3} d (b d+2 c d x)^{3/2}+2 \left (b^2-4 a c\right )^{3/4} d^{5/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \left (b^2-4 a c\right )^{3/4} d^{5/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.77 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {\left (\frac {1}{3}+\frac {i}{3}\right ) (d (b+2 c x))^{5/2} \left ((2-2 i) b \sqrt {b+2 c x}+(4-4 i) c x \sqrt {b+2 c x}-3 \left (b^2-4 a c\right )^{3/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+3 \left (b^2-4 a c\right )^{3/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-3 \left (b^2-4 a c\right )^{3/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{(b+2 c x)^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 139, 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 = {1116, 1118, 27, 25, 266, 827, 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.

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

\(\Big \downarrow \) 1116

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(4*d*(b*d + 2*c*d*x)^(3/2))/3 - 4*(b^2 - 4*a*c)*d^3*(-1/2*ArcTan[Sqrt[b*d 
+ 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/((b^2 - 4*a*c)^(1/4)*Sqrt[d]) + 
ArcTanh[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 4*a*c 
)^(1/4)*Sqrt[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 827

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

rule 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   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] && GtQ[m, 1] && NeQ[m + 2*p 
+ 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
 

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. \(256\) vs. \(2(97)=194\).

Time = 1.54 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.12

method	result	size
derivativedivides	\(4 d \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {d^{2} \left (4 a c -b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(257\)
default	\(4 d \left (\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {d^{2} \left (4 a c -b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(257\)
pseudoelliptic	\(-\frac {d^{2} \left (-8 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \left (2 c x +b \right )+3 \sqrt {2}\, d \left (4 a c -b^{2}\right ) \left (\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )\right )}{6 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\)	\(293\)

Input:

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

Output:

4*d*(1/3*(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)^(1/4) 
*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)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.90 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {4}{3} \, {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d} - \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} + \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) + i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} + i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) - i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} - i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) + \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} - \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) \] Input:

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

Output:

4/3*(2*c*d^2*x + b*d^2)*sqrt(2*c*d*x + b*d) - ((b^6 - 12*a*b^4*c + 48*a^2* 
b^2*c^2 - 64*a^3*c^3)*d^10)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt( 
2*c*d*x + b*d)*d^7 + ((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^1 
0)^(3/4)) + I*((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(1/4 
)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^7 + I*((b^6 - 1 
2*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(3/4)) - I*((b^6 - 12*a*b^4 
*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^ 
2*c^2)*sqrt(2*c*d*x + b*d)*d^7 - I*((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 6 
4*a^3*c^3)*d^10)^(3/4)) + ((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3 
)*d^10)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^7 - 
 ((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(3/4))
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((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. 354 vs. \(2 (97) = 194\).

Time = 0.12 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.93 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=-\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {4}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d \] Input:

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}-2\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4} \] Input:

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

Output:

(4*d*(b*d + 2*c*d*x)^(3/2))/3 + 2*d^(5/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1 
/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(3/4) - 2*d^(5/2)*atanh((b*d + 2*c 
*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(3/4)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.15 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {\sqrt {d}\, d^{2} \left (6 \left (4 a c -b^{2}\right )^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}-2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right )-6 \left (4 a c -b^{2}\right )^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right )-3 \left (4 a c -b^{2}\right )^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right )+3 \left (4 a c -b^{2}\right )^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right )+8 \sqrt {2 c x +b}\, b +16 \sqrt {2 c x +b}\, c x \right )}{6} \] Input:

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

Output:

(sqrt(d)*d**2*(6*(4*a*c - b**2)**(3/4)*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))) - 6*(4*a*c 
- b**2)**(3/4)*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))) - 3*(4*a*c - b**2)**(3/4)*sqrt(2)*l 
og( - sqrt(b + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + 
 b + 2*c*x) + 3*(4*a*c - b**2)**(3/4)*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) + 8*sqrt(b + 2*c*x 
)*b + 16*sqrt(b + 2*c*x)*c*x))/6