\(\int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\) [1288]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {4}{3} d (b d+2 c d x)^{3/2}+2 \left (b^2-4 a c\right )^{3/4} d^{5/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \left (b^2-4 a c\right )^{3/4} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 119, 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 = {706, 708, 335, 304, 209, 212} \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=2 d^{5/2} \left (b^2-4 a c\right )^{3/4} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{5/2} \left (b^2-4 a c\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {4}{3} d (b d+2 c d x)^{3/2} \]

[In]

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

[Out]

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[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 706

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.80 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {\left (\frac {1}{3}+\frac {i}{3}\right ) (d (b+2 c x))^{5/2} \left ((2-2 i) b \sqrt {b+2 c x}+(4-4 i) c x \sqrt {b+2 c x}-3 \left (b^2-4 a c\right )^{3/4} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+3 \left (b^2-4 a c\right )^{3/4} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-3 \left (b^2-4 a c\right )^{3/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{(b+2 c x)^{5/2}} \]

[In]

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

[Out]

((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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(256\) vs. \(2(95)=190\).

Time = 2.68 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.16

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

[In]

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

[Out]

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 = 0.27 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.97 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {4}{3} \, {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d} - \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} + \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) + i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} + i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) - i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} - i \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) + \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} - \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) \]

[In]

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

[Out]

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)*lo
g((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)) + 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 - 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 - 64*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(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((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. 354 vs. \(2 (95) = 190\).

Time = 0.30 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.97 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=-\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {4}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d \]

[In]

integrate((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)^(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 = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx=\frac {4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}-2\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4} \]

[In]

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

[Out]

(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)