### 3.1256 $$\int \frac{1}{(b d+2 c d x) (a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=118 $\frac{16 c^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )^{5/2}}+\frac{8 c}{d \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}$

[Out]

-2/(3*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) + (8*c)/((b^2 - 4*a*c)^2*d*Sqrt[a + b*x + c*x^2]) + (16*c^(3/2)
*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)*d)

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Rubi [A]  time = 0.0767907, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {687, 688, 205} $\frac{16 c^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )^{5/2}}+\frac{8 c}{d \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-2/(3*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) + (8*c)/((b^2 - 4*a*c)^2*d*Sqrt[a + b*x + c*x^2]) + (16*c^(3/2)
*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)*d)

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 688

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
- 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}-\frac{(4 c) \int \frac{1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{3/2}} \, dx}{b^2-4 a c}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{8 c}{\left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{\left (16 c^2\right ) \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{8 c}{\left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{\left (64 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt{a+b x+c x^2}\right )}{\left (b^2-4 a c\right )^2}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{8 c}{\left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{16 c^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d}\\ \end{align*}

Mathematica [C]  time = 0.0360012, size = 62, normalized size = 0.53 $-\frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{3 d \left (b^2-4 a c\right ) (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.192, size = 207, normalized size = 1.8 \begin{align*}{\frac{2}{3\,d \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{-{\frac{3}{2}}}}+8\,{\frac{c}{d \left ( 4\,ac-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{ \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+1/4\,{\frac{4\,ac-{b}^{2}}{c}}}}}}-16\,{\frac{c}{d \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ({ \left ( 1/2\,{\frac{4\,ac-{b}^{2}}{c}}+1/2\,\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}} \right ) \left ( x+1/2\,{\frac{b}{c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d/(4*a*c-b^2)/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+8/d*c/(4*a*c-b^2)^2/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^
2)/c)^(1/2)-16/d*c/(4*a*c-b^2)^2/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1
/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.54834, size = 1341, normalized size = 11.36 \begin{align*} \left [\frac{2 \,{\left (12 \,{\left (c^{3} x^{4} + 2 \, b c^{2} x^{3} + 2 \, a b c x + a^{2} c +{\left (b^{2} c + 2 \, a c^{2}\right )} x^{2}\right )} \sqrt{-\frac{c}{b^{2} - 4 \, a c}} \log \left (-\frac{4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c + 4 \, \sqrt{c x^{2} + b x + a}{\left (b^{2} - 4 \, a c\right )} \sqrt{-\frac{c}{b^{2} - 4 \, a c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) +{\left (12 \, c^{2} x^{2} + 12 \, b c x - b^{2} + 16 \, a c\right )} \sqrt{c x^{2} + b x + a}\right )}}{3 \,{\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d x +{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d\right )}}, \frac{2 \,{\left (24 \,{\left (c^{3} x^{4} + 2 \, b c^{2} x^{3} + 2 \, a b c x + a^{2} c +{\left (b^{2} c + 2 \, a c^{2}\right )} x^{2}\right )} \sqrt{\frac{c}{b^{2} - 4 \, a c}} \arctan \left (-\frac{\sqrt{c x^{2} + b x + a}{\left (b^{2} - 4 \, a c\right )} \sqrt{\frac{c}{b^{2} - 4 \, a c}}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) +{\left (12 \, c^{2} x^{2} + 12 \, b c x - b^{2} + 16 \, a c\right )} \sqrt{c x^{2} + b x + a}\right )}}{3 \,{\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d x +{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[2/3*(12*(c^3*x^4 + 2*b*c^2*x^3 + 2*a*b*c*x + a^2*c + (b^2*c + 2*a*c^2)*x^2)*sqrt(-c/(b^2 - 4*a*c))*log(-(4*c^
2*x^2 + 4*b*c*x - b^2 + 8*a*c + 4*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^2 + 4*b
*c*x + b^2)) + (12*c^2*x^2 + 12*b*c*x - b^2 + 16*a*c)*sqrt(c*x^2 + b*x + a))/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*
c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 -
8*a^2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d), 2/3*(24*(c^3*x^4 + 2*b*c^2*x^3 + 2*
a*b*c*x + a^2*c + (b^2*c + 2*a*c^2)*x^2)*sqrt(c/(b^2 - 4*a*c))*arctan(-1/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)
*sqrt(c/(b^2 - 4*a*c))/(c^2*x^2 + b*c*x + a*c)) + (12*c^2*x^2 + 12*b*c*x - b^2 + 16*a*c)*sqrt(c*x^2 + b*x + a)
)/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*
c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{a^{2} b \sqrt{a + b x + c x^{2}} + 2 a^{2} c x \sqrt{a + b x + c x^{2}} + 2 a b^{2} x \sqrt{a + b x + c x^{2}} + 6 a b c x^{2} \sqrt{a + b x + c x^{2}} + 4 a c^{2} x^{3} \sqrt{a + b x + c x^{2}} + b^{3} x^{2} \sqrt{a + b x + c x^{2}} + 4 b^{2} c x^{3} \sqrt{a + b x + c x^{2}} + 5 b c^{2} x^{4} \sqrt{a + b x + c x^{2}} + 2 c^{3} x^{5} \sqrt{a + b x + c x^{2}}}\, dx}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a**2*b*sqrt(a + b*x + c*x**2) + 2*a**2*c*x*sqrt(a + b*x + c*x**2) + 2*a*b**2*x*sqrt(a + b*x + c*x*
*2) + 6*a*b*c*x**2*sqrt(a + b*x + c*x**2) + 4*a*c**2*x**3*sqrt(a + b*x + c*x**2) + b**3*x**2*sqrt(a + b*x + c*
x**2) + 4*b**2*c*x**3*sqrt(a + b*x + c*x**2) + 5*b*c**2*x**4*sqrt(a + b*x + c*x**2) + 2*c**3*x**5*sqrt(a + b*x
+ c*x**2)), x)/d

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Giac [B]  time = 1.21558, size = 753, normalized size = 6.38 \begin{align*} \frac{32 \, c^{2} \arctan \left (-\frac{2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} c + b \sqrt{c}}{\sqrt{b^{2} c - 4 \, a c^{2}}}\right )}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} \sqrt{b^{2} c - 4 \, a c^{2}}} + \frac{12 \,{\left (\frac{{\left (b^{16} c^{2} d^{3} - 32 \, a b^{14} c^{3} d^{3} + 448 \, a^{2} b^{12} c^{4} d^{3} - 3584 \, a^{3} b^{10} c^{5} d^{3} + 17920 \, a^{4} b^{8} c^{6} d^{3} - 57344 \, a^{5} b^{6} c^{7} d^{3} + 114688 \, a^{6} b^{4} c^{8} d^{3} - 131072 \, a^{7} b^{2} c^{9} d^{3} + 65536 \, a^{8} c^{10} d^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{b^{17} c d^{3} - 32 \, a b^{15} c^{2} d^{3} + 448 \, a^{2} b^{13} c^{3} d^{3} - 3584 \, a^{3} b^{11} c^{4} d^{3} + 17920 \, a^{4} b^{9} c^{5} d^{3} - 57344 \, a^{5} b^{7} c^{6} d^{3} + 114688 \, a^{6} b^{5} c^{7} d^{3} - 131072 \, a^{7} b^{3} c^{8} d^{3} + 65536 \, a^{8} b c^{9} d^{3}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac{b^{18} d^{3} - 48 \, a b^{16} c d^{3} + 960 \, a^{2} b^{14} c^{2} d^{3} - 10752 \, a^{3} b^{12} c^{3} d^{3} + 75264 \, a^{4} b^{10} c^{4} d^{3} - 344064 \, a^{5} b^{8} c^{5} d^{3} + 1032192 \, a^{6} b^{6} c^{6} d^{3} - 1966080 \, a^{7} b^{4} c^{7} d^{3} + 2162688 \, a^{8} b^{2} c^{8} d^{3} - 1048576 \, a^{9} c^{9} d^{3}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\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)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

32*c^2*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^2))/((b^4*d - 8*a*b^2*
c*d + 16*a^2*c^2*d)*sqrt(b^2*c - 4*a*c^2)) + 1/3*(12*((b^16*c^2*d^3 - 32*a*b^14*c^3*d^3 + 448*a^2*b^12*c^4*d^3
- 3584*a^3*b^10*c^5*d^3 + 17920*a^4*b^8*c^6*d^3 - 57344*a^5*b^6*c^7*d^3 + 114688*a^6*b^4*c^8*d^3 - 131072*a^7
*b^2*c^9*d^3 + 65536*a^8*c^10*d^3)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + (b^17*c*d^3 - 32*a*b^15*c^2*d^3 +
448*a^2*b^13*c^3*d^3 - 3584*a^3*b^11*c^4*d^3 + 17920*a^4*b^9*c^5*d^3 - 57344*a^5*b^7*c^6*d^3 + 114688*a^6*b^5*
c^7*d^3 - 131072*a^7*b^3*c^8*d^3 + 65536*a^8*b*c^9*d^3)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x - (b^18*d^3 -
48*a*b^16*c*d^3 + 960*a^2*b^14*c^2*d^3 - 10752*a^3*b^12*c^3*d^3 + 75264*a^4*b^10*c^4*d^3 - 344064*a^5*b^8*c^5*
d^3 + 1032192*a^6*b^6*c^6*d^3 - 1966080*a^7*b^4*c^7*d^3 + 2162688*a^8*b^2*c^8*d^3 - 1048576*a^9*c^9*d^3)/(b^4*
c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2)