3.13.58 \(\int \frac {1}{(b d+2 c d x)^3 (a+b x+c x^2)^{5/2}} \, dx\) [1258]

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

[Out]

-2/3/(-4*a*c+b^2)/d^3/(2*c*x+b)^2/(c*x^2+b*x+a)^(3/2)+40*c^(3/2)*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+
b^2)^(1/2))/(-4*a*c+b^2)^(7/2)/d^3+40/3*c/(-4*a*c+b^2)^2/d^3/(2*c*x+b)^2/(c*x^2+b*x+a)^(1/2)+80*c^2*(c*x^2+b*x
+a)^(1/2)/(-4*a*c+b^2)^3/d^3/(2*c*x+b)^2

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 176, normalized size of antiderivative = 1.00, 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 = {701, 707, 702, 211} \begin {gather*} \frac {40 c^{3/2} \text {ArcTan}\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 {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 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}} \end {gather*}

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 211

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

Rule 701

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 702

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 707

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])

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 ) \text {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 [A]
time = 2.49, size = 169, normalized size = 0.96 \begin {gather*} \frac {\frac {-2 b^4+40 b^3 c x+160 b c^2 x \left (2 a+3 c x^2\right )+56 b^2 c \left (a+5 c x^2\right )+16 c^2 \left (3 a^2+20 a c x^2+15 c^2 x^4\right )}{3 \left (b^2-4 a c\right )^3 (b+2 c x)^2 (a+x (b+c x))^{3/2}}-\frac {80 c^{3/2} \tan ^{-1}\left (\frac {b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}}{d^3} \end {gather*}

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*b^4 + 40*b^3*c*x + 160*b*c^2*x*(2*a + 3*c*x^2) + 56*b^2*c*(a + 5*c*x^2) + 16*c^2*(3*a^2 + 20*a*c*x^2 + 15
*c^2*x^4))/(3*(b^2 - 4*a*c)^3*(b + 2*c*x)^2*(a + x*(b + c*x))^(3/2)) - (80*c^(3/2)*ArcTan[(b + 2*c*x - 2*Sqrt[
c]*Sqrt[a + x*(b + c*x)])/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(7/2))/d^3

________________________________________________________________________________________

Maple [A]
time = 0.72, size = 296, normalized size = 1.68

method	result	size
default	\(\frac {-\frac {2 c}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2} \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}-\frac {10 c^{2} \left (\frac {4 c}{3 \left (4 a c -b^{2}\right ) \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}+\frac {4 c \left (\frac {4 c}{\left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}-\frac {8 c \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}\right )}{4 a c -b^{2}}}{8 d^{3} c^{3}}\)	\(296\)

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,method=_RETURNVERBOSE)

[Out]

1/8/d^3/c^3*(-2/(4*a*c-b^2)*c/(x+1/2*b/c)^2/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)-10*c^2/(4*a*c-b^2)*(4/3/
(4*a*c-b^2)*c/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+4/(4*a*c-b^2)*c*(4/(4*a*c-b^2)*c/((x+1/2*b/c)^2*c+1/4*
(4*a*c-b^2)/c)^(1/2)-8/(4*a*c-b^2)*c/((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)))))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 628 vs. \(2 (156) = 312\).
time = 8.82, size = 1287, normalized size = 7.31 \begin {gather*} \left [-\frac {2 \, {\left (30 \, {\left (4 \, c^{5} x^{6} + 12 \, b c^{4} x^{5} + a^{2} b^{2} c + {\left (13 \, b^{2} c^{3} + 8 \, a c^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} x^{3} + {\left (b^{4} c + 10 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} x^{2} + 2 \, {\left (a b^{3} c + 2 \, a^{2} b c^{2}\right )} x\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 (120 \, c^{4} x^{4} + 240 \, b c^{3} x^{3} - b^{4} + 28 \, a b^{2} c + 24 \, a^{2} c^{2} + 20 \, {\left (7 \, b^{2} c^{2} + 8 \, a c^{3}\right )} x^{2} + 20 \, {\left (b^{3} c + 8 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (4 \, {\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{3} x^{6} + 12 \, {\left (b^{7} c^{3} - 12 \, a b^{5} c^{4} + 48 \, a^{2} b^{3} c^{5} - 64 \, a^{3} b c^{6}\right )} d^{3} x^{5} + {\left (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}\right )} d^{3} x^{4} + 2 \, {\left (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}\right )} d^{3} x^{3} + {\left (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}\right )} d^{3} x^{2} + 2 \, {\left (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}\right )} d^{3} x + {\left (a^{2} b^{8} - 12 \, a^{3} b^{6} c + 48 \, a^{4} b^{4} c^{2} - 64 \, a^{5} b^{2} c^{3}\right )} d^{3}\right )}}, \frac {2 \, {\left (60 \, {\left (4 \, c^{5} x^{6} + 12 \, b c^{4} x^{5} + a^{2} b^{2} c + {\left (13 \, b^{2} c^{3} + 8 \, a c^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} x^{3} + {\left (b^{4} c + 10 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} x^{2} + 2 \, {\left (a b^{3} c + 2 \, a^{2} b c^{2}\right )} x\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 (120 \, c^{4} x^{4} + 240 \, b c^{3} x^{3} - b^{4} + 28 \, a b^{2} c + 24 \, a^{2} c^{2} + 20 \, {\left (7 \, b^{2} c^{2} + 8 \, a c^{3}\right )} x^{2} + 20 \, {\left (b^{3} c + 8 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (4 \, {\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{3} x^{6} + 12 \, {\left (b^{7} c^{3} - 12 \, a b^{5} c^{4} + 48 \, a^{2} b^{3} c^{5} - 64 \, a^{3} b c^{6}\right )} d^{3} x^{5} + {\left (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}\right )} d^{3} x^{4} + 2 \, {\left (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}\right )} d^{3} x^{3} + {\left (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}\right )} d^{3} x^{2} + 2 \, {\left (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}\right )} d^{3} x + {\left (a^{2} b^{8} - 12 \, a^{3} b^{6} c + 48 \, a^{4} b^{4} c^{2} - 64 \, a^{5} b^{2} c^{3}\right )} d^{3}\right )}}\right ] \end {gather*}

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)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.43Unable to divide, perhaps due to rounding error%%%{%%%{512,[6]%%%},[6,3,3,0]%%%}+%%%{%
%%{-384,[5]

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^3\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________