3.1292 \(\int \frac{1}{(b d+2 c d x)^{3/2} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=129 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4}}+\frac{4}{d \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}} \]

[Out]

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

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Rubi [A]  time = 0.108493, antiderivative size = 129, 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 = {693, 694, 329, 298, 203, 206} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4}}+\frac{4}{d \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)),x]

[Out]

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

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 298

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 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])

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])

Rubi steps

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

Mathematica [C]  time = 0.0447679, size = 54, normalized size = 0.42 \[ \frac{4 \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right ) \sqrt{d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)),x]

[Out]

(4*Hypergeometric2F1[-1/4, 1, 3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)*d*Sqrt[d*(b + 2*c*x)])

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Maple [B]  time = 0.195, size = 341, normalized size = 2.6 \begin{align*} -{\frac{\sqrt{2}}{2\,d \left ( 4\,ac-{b}^{2} \right ) }\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 ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}-{\frac{\sqrt{2}}{d \left ( 4\,ac-{b}^{2} \right ) }\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+{\frac{\sqrt{2}}{d \left ( 4\,ac-{b}^{2} \right ) }\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}-4\,{\frac{1}{d \left ( 4\,ac-{b}^{2} \right ) \sqrt{2\,cdx+bd}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x)

[Out]

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

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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(1/(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 = 2.38624, size = 1912, normalized size = 14.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

(4*(2*(b^2*c - 4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*d^2)*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^
3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6))^(1/4)*arctan(-sqrt((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)
*d^4*sqrt(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6)) +
2*c*d*x + b*d)*(b^2 - 4*a*c)*d*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4
- 1024*a^5*c^5)*d^6))^(1/4) + sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)*d*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 6
40*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6))^(1/4)) - (2*(b^2*c - 4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*
d^2)*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6))^(1/4)
*log((b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*d^5*(1/((b^10 - 20*a*b^8*c + 160*a^2*
b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6))^(3/4) + sqrt(2*c*d*x + b*d)) + (2*(b^2*c -
4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*d^2)*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^
2*c^4 - 1024*a^5*c^5)*d^6))^(1/4)*log(-(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*d^5
*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6))^(3/4) + s
qrt(2*c*d*x + b*d)) + 4*sqrt(2*c*d*x + b*d))/(2*(b^2*c - 4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \left (b + 2 c x\right )\right )^{\frac{3}{2}} \left (a + b x + c x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a),x)

[Out]

Integral(1/((d*(b + 2*c*x))**(3/2)*(a + b*x + c*x**2)), x)

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Giac [B]  time = 1.15078, size = 671, normalized size = 5.2 \begin{align*} -\frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \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 )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \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 )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} + \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \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 )}{\sqrt{2} b^{4} d^{3} - 8 \, \sqrt{2} a b^{2} c d^{3} + 16 \, \sqrt{2} a^{2} c^{2} d^{3}} - \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \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 )}{\sqrt{2} b^{4} d^{3} - 8 \, \sqrt{2} a b^{2} c d^{3} + 16 \, \sqrt{2} a^{2} c^{2} d^{3}} + \frac{4}{{\left (b^{2} d - 4 \, a c d\right )} \sqrt{2 \, c d x + b d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)^(3/4)*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))/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) - sqrt(2)*(-b^2*d^2 + 4*a*c
*d^2)^(3/4)*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))/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) + (-b^2*d^2 + 4*a*c*d^2)^(3/4)*log(2*c*d*x + b*d + s
qrt(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))/(sqrt(2)*b^4*d^3 - 8*sqr
t(2)*a*b^2*c*d^3 + 16*sqrt(2)*a^2*c^2*d^3) - (-b^2*d^2 + 4*a*c*d^2)^(3/4)*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))/(sqrt(2)*b^4*d^3 - 8*sqrt(2)*a*b^2*c*d^
3 + 16*sqrt(2)*a^2*c^2*d^3) + 4/((b^2*d - 4*a*c*d)*sqrt(2*c*d*x + b*d))