\(\int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\) [1289]

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=4 d \sqrt {b d+2 c d x}-2 \sqrt [4]{b^2-4 a c} d^{3/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \sqrt [4]{b^2-4 a c} d^{3/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 117, 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, 218, 212, 209} \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=-2 d^{3/2} \sqrt [4]{b^2-4 a c} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{3/2} \sqrt [4]{b^2-4 a c} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+4 d \sqrt {b d+2 c d x} \]

[In]

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

[Out]

4*d*Sqrt[b*d + 2*c*d*x] - 2*(b^2 - 4*a*c)^(1/4)*d^(3/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d
])] - 2*(b^2 - 4*a*c)^(1/4)*d^(3/2)*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 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 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}& = 4 d \sqrt {b d+2 c d x}+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx \\ & = 4 d \sqrt {b d+2 c d x}+\frac {\left (\left (b^2-4 a c\right ) d\right ) \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} \\ & = 4 d \sqrt {b d+2 c d x}+\frac {\left (\left (b^2-4 a c\right ) d\right ) \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} \\ & = 4 d \sqrt {b d+2 c d x}-\left (2 \sqrt {b^2-4 a c} d^2\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 \sqrt {b^2-4 a c} d^2\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 ) \\ & = 4 d \sqrt {b d+2 c d x}-2 \sqrt [4]{b^2-4 a c} d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \sqrt [4]{b^2-4 a c} d^{3/2} \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 = 0.19 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.64 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=\frac {(1-i) (d (b+2 c x))^{3/2} \left ((2+2 i) \sqrt {b+2 c x}+\sqrt [4]{b^2-4 a c} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\sqrt [4]{b^2-4 a c} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\sqrt [4]{b^2-4 a c} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{(b+2 c x)^{3/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(95)=190\).

Time = 2.65 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.18

method	result	size
derivativedivides	\(4 d \left (\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(255\)
default	\(4 d \left (\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(255\)
pseudoelliptic	\(-\frac {2 d \left (-2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \sqrt {d \left (2 c x +b \right )}+\frac {\sqrt {2}\, d^{2} \left (4 a c -b^{2}\right ) \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{4}\right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}\)	\(286\)

[In]

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

[Out]

4*d*((2*c*d*x+b*d)^(1/2)-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*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 = 200, normalized size of antiderivative = 1.71 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=4 \, \sqrt {2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d + \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) - i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d + i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) + i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d - i \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) + \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((2*c*d*x+b*d)^(3/2)/(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 = 0.29 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.03 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=-\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + 4 \, \sqrt {2 \, c d x + b d} d \]

[In]

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

[Out]

-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 = 9.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.83 \[ \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx=4\,d\,\sqrt {b\,d+2\,c\,d\,x}-2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-2\,d^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4} \]

[In]

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

[Out]

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)