3.1289 \(\int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=117 \[ -2 d^{3/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{3/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )+4 d \sqrt{b d+2 c d x} \]

[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])]

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Rubi [A]  time = 0.0963116, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {692, 694, 329, 212, 206, 203} \[ -2 d^{3/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{3/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )+4 d \sqrt{b d+2 c d x} \]

Antiderivative was successfully verified.

[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 692

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 694

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]

Rule 329

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 212

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] &&
!GtQ[a/b, 0]

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \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}+\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 [A]  time = 0.0476966, size = 113, normalized size = 0.97 \[ \frac{2 (d (b+2 c x))^{3/2} \left (-\sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+2 \sqrt{b+2 c x}\right )}{(b+2 c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.19, size = 582, normalized size = 5. \begin{align*} 4\,d\sqrt{2\,cdx+bd}+4\,{\frac{{d}^{3}\sqrt{2}ac}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }-{{d}^{3}\sqrt{2}{b}^{2}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-2\,{\frac{{d}^{3}\sqrt{2}ac}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\ln \left ({\frac{2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}{2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}} \right ) }+{\frac{{d}^{3}\sqrt{2}{b}^{2}}{2}\ln \left ({ \left ( 2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) \left ( 2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) ^{-1}} \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-4\,{\frac{{d}^{3}\sqrt{2}ac}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }+{{d}^{3}\sqrt{2}{b}^{2}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.8926, size = 509, normalized size = 4.35 \begin{align*} 4 \, \sqrt{2 \, c d x + b d} d - 4 \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac{1}{4}} \arctan \left (\frac{\left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac{3}{4}} \sqrt{2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac{3}{4}} \sqrt{2 \, c d^{3} x + b d^{3} + \sqrt{{\left (b^{2} - 4 \, a c\right )} d^{6}}}}{{\left (b^{2} - 4 \, a c\right )} d^{6}}\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 ) + \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 - 4*((b^2 - 4*a*c)*d^6)^(1/4)*arctan((((b^2 - 4*a*c)*d^6)^(3/4)*sqrt(2*c*d*x + b*d)*d
- ((b^2 - 4*a*c)*d^6)^(3/4)*sqrt(2*c*d^3*x + b*d^3 + sqrt((b^2 - 4*a*c)*d^6)))/((b^2 - 4*a*c)*d^6)) - ((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)) + ((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))

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Sympy [A]  time = 37.3032, size = 212, normalized size = 1.81 \begin{align*} - 16 a c d^{3} \operatorname{RootSum}{\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left ( t \mapsto t \log{\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt{b d + 2 c d x} \right )} \right )\right )} + 4 b^{2} d^{3} \operatorname{RootSum}{\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left ( t \mapsto t \log{\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt{b d + 2 c d x} \right )} \right )\right )} + 4 d \sqrt{b d + 2 c d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*a*c*d**3*RootSum(_t**4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d
**6) + 1, Lambda(_t, _t*log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 4*b**2*d**3*RootSum(_t*
*4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d**6) + 1, Lambda(_t, _t*
log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 4*d*sqrt(b*d + 2*c*d*x)

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Giac [B]  time = 1.23847, size = 478, normalized size = 4.09 \begin{align*} -\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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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