\(\int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\) [113]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=4 d \sqrt {b d+2 c d x}-2 \sqrt [4]{b^2-4 a c} d^{3/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \sqrt [4]{b^2-4 a c} d^{3/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \] Output:

4*d*(2*c*d*x+b*d)^(1/2)-2*(-4*a*c+b^2)^(1/4)*d^(3/2)*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)^(1/4)*d^(3/2)*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 = 0.23 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.61 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=\frac {(1-i) (d (b+2 c x))^{3/2} \left ((2+2 i) \sqrt {b+2 c x}+\sqrt [4]{b^2-4 a c} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\sqrt [4]{b^2-4 a c} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\sqrt [4]{b^2-4 a c} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{(b+2 c x)^{3/2}} \] Input:

Integrate[(b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 137, 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, 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.

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

\(\Big \downarrow \) 1116

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[(b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2),x]
 

Output:

4*d*Sqrt[b*d + 2*c*d*x] - 4*(b^2 - 4*a*c)*d^3*(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)) + ArcTanh[S 
qrt[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)))
 

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 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. \(252\) vs. \(2(97)=194\).

Time = 1.43 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.13

method	result	size
pseudoelliptic	\(-\frac {2 d \left (-2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \sqrt {d \left (2 c x +b \right )}+\frac {d^{2} \sqrt {2}\, \left (4 a c -b^{2}\right ) \left (\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right )\right )}{4}\right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}\)	\(253\)
derivativedivides	\(4 d \left (\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(255\)
default	\(4 d \left (\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(255\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.68 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=4 \, \sqrt {2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d + \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) - i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d + i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) + i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d - i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) + \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) \] Input:

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x, algorithm="fricas")
 

Output:

4*sqrt(2*c*d*x + b*d)*d - ((b^2 - 4*a*c)*d^6)^(1/4)*log(sqrt(2*c*d*x + b*d 
)*d + ((b^2 - 4*a*c)*d^6)^(1/4)) - I*((b^2 - 4*a*c)*d^6)^(1/4)*log(sqrt(2* 
c*d*x + b*d)*d + I*((b^2 - 4*a*c)*d^6)^(1/4)) + I*((b^2 - 4*a*c)*d^6)^(1/4 
)*log(sqrt(2*c*d*x + b*d)*d - I*((b^2 - 4*a*c)*d^6)^(1/4)) + ((b^2 - 4*a*c 
)*d^6)^(1/4)*log(sqrt(2*c*d*x + b*d)*d - ((b^2 - 4*a*c)*d^6)^(1/4))
 

Sympy [F]

\[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}}}{a + b x + c x^{2}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*d*x+b*d)^(3/2)/(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 = 0.13 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.97 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=-\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + 4 \, \sqrt {2 \, c d x + b d} d \] Input:

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=4\,d\,\sqrt {b\,d+2\,c\,d\,x}-2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-2\,d^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.06 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=\frac {\sqrt {d}\, d \left (2 \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}-2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right )-2 \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+2 \sqrt {2 c x +b}}{\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}}\right )+\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right )-\left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {2 c x +b}\, \left (4 a c -b^{2}\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {4 a c -b^{2}}+b +2 c x \right )+8 \sqrt {2 c x +b}\right )}{2} \] Input:

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

Output:

(sqrt(d)*d*(2*(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))) - 2*(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))) + (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) - (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) + 8*sqrt(b + 2*c*x)))/2