3.1258 \(\int \frac{1}{(b d+2 c d x)^3 (a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=176 \[ \frac{80 c^2 \sqrt{a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^3 (b+2 c x)^2}+\frac{40 c^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{7/2}}+\frac{40 c}{3 d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{2}{3 d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

-2/(3*(b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(3/2)) + (40*c)/(3*(b^2 - 4*a*c)^2*d^3*(b + 2*c*x)^2*S
qrt[a + b*x + c*x^2]) + (80*c^2*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^3*d^3*(b + 2*c*x)^2) + (40*c^(3/2)*ArcTa
n[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(7/2)*d^3)

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Rubi [A]  time = 0.105792, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {687, 693, 688, 205} \[ \frac{80 c^2 \sqrt{a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^3 (b+2 c x)^2}+\frac{40 c^{3/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{7/2}}+\frac{40 c}{3 d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2 \sqrt{a+b x+c x^2}}-\frac{2}{3 d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2)),x]

[Out]

-2/(3*(b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(3/2)) + (40*c)/(3*(b^2 - 4*a*c)^2*d^3*(b + 2*c*x)^2*S
qrt[a + b*x + c*x^2]) + (80*c^2*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^3*d^3*(b + 2*c*x)^2) + (40*c^(3/2)*ArcTa
n[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(7/2)*d^3)

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 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 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)^3 \left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{(20 c) \int \frac{1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac{40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}+\frac{\left (80 c^2\right ) \int \frac{1}{(b d+2 c d x)^3 \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^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac{40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}+\frac{80 c^2 \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac{\left (40 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 )^3 d^2}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac{40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}+\frac{80 c^2 \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac{\left (160 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 )^3 d^2}\\ &=-\frac{2}{3 \left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}}+\frac{40 c}{3 \left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2 \sqrt{a+b x+c x^2}}+\frac{80 c^2 \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^3 d^3 (b+2 c x)^2}+\frac{40 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 )^{7/2} d^3}\\ \end{align*}

Mathematica [C]  time = 0.0364234, size = 62, normalized size = 0.35 \[ -\frac{2 \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{3 d^3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.2, size = 267, normalized size = 1.5 \begin{align*} -{\frac{1}{4\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) } \left ( x+{\frac{b}{2\,c}} \right ) ^{-2} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{-{\frac{3}{2}}}}-{\frac{5}{3\,{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{-{\frac{3}{2}}}}-20\,{\frac{c}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}}{\frac{1}{\sqrt{ \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+1/4\,{\frac{4\,ac-{b}^{2}}{c}}}}}}+40\,{\frac{c}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}}\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)^3/(c*x^2+b*x+a)^(5/2),x)

[Out]

-1/4/d^3/c^2/(4*a*c-b^2)/(x+1/2*b/c)^2/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)-5/3/d^3/(4*a*c-b^2)^2/((x+1/2
*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)-20/d^3*c/(4*a*c-b^2)^3/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)+40/d^3*c/(
4*a*c-b^2)^3/((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)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 40.8697, size = 2759, normalized size = 15.68 \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/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-2/3*(30*(4*c^5*x^6 + 12*b*c^4*x^5 + a^2*b^2*c + (13*b^2*c^3 + 8*a*c^4)*x^4 + 2*(3*b^3*c^2 + 8*a*b*c^3)*x^3 +
(b^4*c + 10*a*b^2*c^2 + 4*a^2*c^3)*x^2 + 2*(a*b^3*c + 2*a^2*b*c^2)*x)*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)) - (120*c^4*x^4 + 240*b*c^3*x^3 - b^4 + 28*a*b^2*c + 24*a^2*c^2 + 20*(7*b^2*c^2 + 8*a*c^3)*x^2 + 20*(b^3
*c + 8*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^3*x^6 +
12*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 + 528*a^2*b^
4*c^4 - 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^3*b^3*c^4
- 512*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 - 256*a^5*c^
5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2*b^8 - 12*
a^3*b^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3), 2/3*(60*(4*c^5*x^6 + 12*b*c^4*x^5 + a^2*b^2*c + (13*b^2*c^3
+ 8*a*c^4)*x^4 + 2*(3*b^3*c^2 + 8*a*b*c^3)*x^3 + (b^4*c + 10*a*b^2*c^2 + 4*a^2*c^3)*x^2 + 2*(a*b^3*c + 2*a^2*
b*c^2)*x)*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)) + (120*c^4*x^4 + 240*b*c^3*x^3 - b^4 + 28*a*b^2*c + 24*a^2*c^2 + 20*(7*b^2*c^2 + 8*a*c^3)*x^2
+ 20*(b^3*c + 8*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*
d^3*x^6 + 12*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 +
528*a^2*b^4*c^4 - 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^
3*b^3*c^4 - 512*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 -
256*a^5*c^5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2
*b^8 - 12*a^3*b^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{a^{2} b^{3} \sqrt{a + b x + c x^{2}} + 6 a^{2} b^{2} c x \sqrt{a + b x + c x^{2}} + 12 a^{2} b c^{2} x^{2} \sqrt{a + b x + c x^{2}} + 8 a^{2} c^{3} x^{3} \sqrt{a + b x + c x^{2}} + 2 a b^{4} x \sqrt{a + b x + c x^{2}} + 14 a b^{3} c x^{2} \sqrt{a + b x + c x^{2}} + 36 a b^{2} c^{2} x^{3} \sqrt{a + b x + c x^{2}} + 40 a b c^{3} x^{4} \sqrt{a + b x + c x^{2}} + 16 a c^{4} x^{5} \sqrt{a + b x + c x^{2}} + b^{5} x^{2} \sqrt{a + b x + c x^{2}} + 8 b^{4} c x^{3} \sqrt{a + b x + c x^{2}} + 25 b^{3} c^{2} x^{4} \sqrt{a + b x + c x^{2}} + 38 b^{2} c^{3} x^{5} \sqrt{a + b x + c x^{2}} + 28 b c^{4} x^{6} \sqrt{a + b x + c x^{2}} + 8 c^{5} x^{7} \sqrt{a + b x + c x^{2}}}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral(1/(a**2*b**3*sqrt(a + b*x + c*x**2) + 6*a**2*b**2*c*x*sqrt(a + b*x + c*x**2) + 12*a**2*b*c**2*x**2*sq
rt(a + b*x + c*x**2) + 8*a**2*c**3*x**3*sqrt(a + b*x + c*x**2) + 2*a*b**4*x*sqrt(a + b*x + c*x**2) + 14*a*b**3
*c*x**2*sqrt(a + b*x + c*x**2) + 36*a*b**2*c**2*x**3*sqrt(a + b*x + c*x**2) + 40*a*b*c**3*x**4*sqrt(a + b*x +
c*x**2) + 16*a*c**4*x**5*sqrt(a + b*x + c*x**2) + b**5*x**2*sqrt(a + b*x + c*x**2) + 8*b**4*c*x**3*sqrt(a + b*
x + c*x**2) + 25*b**3*c**2*x**4*sqrt(a + b*x + c*x**2) + 38*b**2*c**3*x**5*sqrt(a + b*x + c*x**2) + 28*b*c**4*
x**6*sqrt(a + b*x + c*x**2) + 8*c**5*x**7*sqrt(a + b*x + c*x**2)), x)/d**3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError