### 3.1293 $$\int \frac{1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=131 $-\frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}+\frac{4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}$

[Out]

4/(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) - (2*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((
b^2 - 4*a*c)^(7/4)*d^(5/2)) - (2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(7
/4)*d^(5/2))

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Rubi [A]  time = 0.107208, antiderivative size = 131, 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, 212, 206, 203} $-\frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{7/4}}+\frac{4}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

4/(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) - (2*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((
b^2 - 4*a*c)^(7/4)*d^(5/2)) - (2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(7
/4)*d^(5/2))

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 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{1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx &=\frac{4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac{\int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) d^2}\\ &=\frac{4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac{\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 \left (b^2-4 a c\right ) d^3}\\ &=\frac{4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac{\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 \left (b^2-4 a c\right ) d^3}\\ &=\frac{4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}-\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 )^{3/2} d^2}-\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 )^{3/2} d^2}\\ &=\frac{4}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}-\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 )^{7/4} d^{5/2}}-\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 )^{7/4} d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0528148, size = 56, normalized size = 0.43 $\frac{4 \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right ) (d (b+2 c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)),x]

[Out]

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

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Maple [B]  time = 0.192, size = 341, normalized size = 2.6 \begin{align*} -{\frac{4}{3\,d \left ( 4\,ac-{b}^{2} \right ) } \left ( 2\,cdx+bd \right ) ^{-{\frac{3}{2}}}}-{\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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-{\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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}+{\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 ) \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(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.34273, size = 2742, normalized size = 20.93 \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)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

-1/3*(12*(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*a*b^2*c)*d^3)*(1/((b^14 - 28*
a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16
384*a^7*c^7)*d^10))^(1/4)*arctan(-(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)*sqrt((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^6*sqrt(1/((b^14 - 28*
a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16
384*a^7*c^7)*d^10)) + 2*c*d*x + b*d)*d^7*(1/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*
a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^10))^(3/4) + (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)*sqrt(2*c*d*x + b*d)*d^7*(1/((b^14 - 28*a*b^12*
c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7
*c^7)*d^10))^(3/4)) + 3*(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*a*b^2*c)*d^3)*
(1/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a
^6*b^2*c^6 - 16384*a^7*c^7)*d^10))^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*(1/((b^14 - 28*a*b^12*c + 336*
a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^
10))^(1/4) + sqrt(2*c*d*x + b*d)) - 3*(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*
a*b^2*c)*d^3)*(1/((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4
*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^10))^(1/4)*log(-(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*(1/((b^14 - 28*
a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16
384*a^7*c^7)*d^10))^(1/4) + sqrt(2*c*d*x + b*d)) - 4*sqrt(2*c*d*x + b*d))/(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(
b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4*a*b^2*c)*d^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 1.1978, size = 671, normalized size = 5.12 \begin{align*} -\frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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{1}{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{1}{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{1}{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}{3 \,{\left (b^{2} d - 4 \, a c d\right )}{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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