3.13.57 \(\int \frac {1}{(b d+2 c d x)^2 (a+b x+c x^2)^{5/2}} \, dx\) [1257]

Optimal. Leaf size=122 \[ -\frac {2}{3 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}+\frac {32 c}{3 \left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}+\frac {256 c^2 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right )^3 d^2 (b+2 c x)} \]

[Out]

-2/3/(-4*a*c+b^2)/d^2/(2*c*x+b)/(c*x^2+b*x+a)^(3/2)+32/3*c/(-4*a*c+b^2)^2/d^2/(2*c*x+b)/(c*x^2+b*x+a)^(1/2)+25
6/3*c^2*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^3/d^2/(2*c*x+b)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {701, 696} \begin {gather*} \frac {256 c^2 \sqrt {a+b x+c x^2}}{3 d^2 \left (b^2-4 a c\right )^3 (b+2 c x)}+\frac {32 c}{3 d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}-\frac {2}{3 d^2 \left (b^2-4 a c\right ) (b+2 c x) \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)^2*(a + b*x + c*x^2)^(5/2)),x]

[Out]

-2/(3*(b^2 - 4*a*c)*d^2*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2)) + (32*c)/(3*(b^2 - 4*a*c)^2*d^2*(b + 2*c*x)*Sqrt[
a + b*x + c*x^2]) + (256*c^2*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*c)^3*d^2*(b + 2*c*x))

Rule 696

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] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]

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]

Rubi steps

\begin {align*} \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2}{3 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}-\frac {(16 c) \int \frac {1}{(b d+2 c d x)^2 \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^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}+\frac {32 c}{3 \left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}+\frac {\left (128 c^2\right ) \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2}\\ &=-\frac {2}{3 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}+\frac {32 c}{3 \left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}+\frac {256 c^2 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right )^3 d^2 (b+2 c x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.56, size = 108, normalized size = 0.89 \begin {gather*} \frac {-2 b^4+32 b^3 c x+128 b c^2 x \left (3 a+4 c x^2\right )+48 b^2 c \left (a+6 c x^2\right )+32 c^2 \left (3 a^2+12 a c x^2+8 c^2 x^4\right )}{3 \left (b^2-4 a c\right )^3 d^2 (b+2 c x) (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^4 + 32*b^3*c*x + 128*b*c^2*x*(3*a + 4*c*x^2) + 48*b^2*c*(a + 6*c*x^2) + 32*c^2*(3*a^2 + 12*a*c*x^2 + 8*c
^2*x^4))/(3*(b^2 - 4*a*c)^3*d^2*(b + 2*c*x)*(a + x*(b + c*x))^(3/2))

________________________________________________________________________________________

Maple [A]
time = 0.71, size = 191, normalized size = 1.57

method	result	size
trager	\(-\frac {2 \left (128 c^{4} x^{4}+256 b \,c^{3} x^{3}+192 x^{2} c^{3} a +144 b^{2} c^{2} x^{2}+192 x a b \,c^{2}+16 b^{3} c x +48 a^{2} c^{2}+24 a c \,b^{2}-b^{4}\right )}{3 d^{2} \left (2 c x +b \right ) \left (16 a^{2} c^{2}-8 a c \,b^{2}+b^{4}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )}\)	\(132\)
gosper	\(-\frac {2 \left (128 c^{4} x^{4}+256 b \,c^{3} x^{3}+192 x^{2} c^{3} a +144 b^{2} c^{2} x^{2}+192 x a b \,c^{2}+16 b^{3} c x +48 a^{2} c^{2}+24 a c \,b^{2}-b^{4}\right )}{3 \left (2 c x +b \right ) d^{2} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(133\)
default	\(\frac {-\frac {4 c}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right ) \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}-\frac {16 c^{2} \left (\frac {4 c \left (x +\frac {b}{2 c}\right )}{3 \left (4 a c -b^{2}\right ) \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}+\frac {32 c^{2} \left (x +\frac {b}{2 c}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}\right )}{4 a c -b^{2}}}{4 d^{2} c^{2}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d^2/c^2*(-4/(4*a*c-b^2)*c/(x+1/2*b/c)/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)-16*c^2/(4*a*c-b^2)*(4/3*c*
(x+1/2*b/c)/(4*a*c-b^2)/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+32/3*c^2/(4*a*c-b^2)^2*(x+1/2*b/c)/((x+1/2*b
/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)))

________________________________________________________________________________________

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)^2/(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. 374 vs. \(2 (110) = 220\).
time = 5.91, size = 374, normalized size = 3.07 \begin {gather*} \frac {2 \, {\left (128 \, c^{4} x^{4} + 256 \, b c^{3} x^{3} - b^{4} + 24 \, a b^{2} c + 48 \, a^{2} c^{2} + 48 \, {\left (3 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} + 16 \, {\left (b^{3} c + 12 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (2 \, {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} d^{2} x^{5} + 5 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} d^{2} x^{4} + 4 \, {\left (b^{8} c - 11 \, a b^{6} c^{2} + 36 \, a^{2} b^{4} c^{3} - 16 \, a^{3} b^{2} c^{4} - 64 \, a^{4} c^{5}\right )} d^{2} x^{3} + {\left (b^{9} - 6 \, a b^{7} c - 24 \, a^{2} b^{5} c^{2} + 224 \, a^{3} b^{3} c^{3} - 384 \, a^{4} b c^{4}\right )} d^{2} x^{2} + 2 \, {\left (a b^{8} - 11 \, a^{2} b^{6} c + 36 \, a^{3} b^{4} c^{2} - 16 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4}\right )} d^{2} x + {\left (a^{2} b^{7} - 12 \, a^{3} b^{5} c + 48 \, a^{4} b^{3} c^{2} - 64 \, a^{5} b c^{3}\right )} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(128*c^4*x^4 + 256*b*c^3*x^3 - b^4 + 24*a*b^2*c + 48*a^2*c^2 + 48*(3*b^2*c^2 + 4*a*c^3)*x^2 + 16*(b^3*c +
12*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a)/(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^
7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*
a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x
^2 + 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c +
48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{a^{2} b^{2} \sqrt {a + b x + c x^{2}} + 4 a^{2} b c x \sqrt {a + b x + c x^{2}} + 4 a^{2} c^{2} x^{2} \sqrt {a + b x + c x^{2}} + 2 a b^{3} x \sqrt {a + b x + c x^{2}} + 10 a b^{2} c x^{2} \sqrt {a + b x + c x^{2}} + 16 a b c^{2} x^{3} \sqrt {a + b x + c x^{2}} + 8 a c^{3} x^{4} \sqrt {a + b x + c x^{2}} + b^{4} x^{2} \sqrt {a + b x + c x^{2}} + 6 b^{3} c x^{3} \sqrt {a + b x + c x^{2}} + 13 b^{2} c^{2} x^{4} \sqrt {a + b x + c x^{2}} + 12 b c^{3} x^{5} \sqrt {a + b x + c x^{2}} + 4 c^{4} x^{6} \sqrt {a + b x + c x^{2}}}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a**2*b**2*sqrt(a + b*x + c*x**2) + 4*a**2*b*c*x*sqrt(a + b*x + c*x**2) + 4*a**2*c**2*x**2*sqrt(a +
b*x + c*x**2) + 2*a*b**3*x*sqrt(a + b*x + c*x**2) + 10*a*b**2*c*x**2*sqrt(a + b*x + c*x**2) + 16*a*b*c**2*x**
3*sqrt(a + b*x + c*x**2) + 8*a*c**3*x**4*sqrt(a + b*x + c*x**2) + b**4*x**2*sqrt(a + b*x + c*x**2) + 6*b**3*c*
x**3*sqrt(a + b*x + c*x**2) + 13*b**2*c**2*x**4*sqrt(a + b*x + c*x**2) + 12*b*c**3*x**5*sqrt(a + b*x + c*x**2)
+ 4*c**4*x**6*sqrt(a + b*x + c*x**2)), x)/d**2

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (110) = 220\).
time = 1.59, size = 374, normalized size = 3.07 \begin {gather*} \frac {16 \, {\left (\frac {{\left (\frac {3 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} c^{2}}{b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )} + \frac {6 \, {\left (\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - c\right )} c^{3} + c^{4}}{{\left (b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )\right )} {\left (\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - c\right )} \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}\right )} c^{5} d^{8} {\left | c \right |}}{{\left (b^{2} c^{3} d^{4} - 4 \, a c^{4} d^{4}\right )}^{2}} - \frac {8 \, c^{\frac {3}{2}} {\left | c \right |} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )}}{3 \, d^{2} {\left | c \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*((3*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*c^2/(b^2*sgn(1/(2*c*d*x + b*d)
)*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)) + (6*(b^2*c*d^2/(2*c*d*x + b*d)^2 - 4*a*c^2*d^2/
(2*c*d*x + b*d)^2 - c)*c^3 + c^4)/((b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sg
n(c)*sgn(d))*(b^2*c*d^2/(2*c*d*x + b*d)^2 - 4*a*c^2*d^2/(2*c*d*x + b*d)^2 - c)*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)
^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)))*c^5*d^8*abs(c)/(b^2*c^3*d^4 - 4*a*c^4*d^4)^2 - 8*c^(3/2)*abs(c)*sgn(
1/(2*c*d*x + b*d))*sgn(c)*sgn(d)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))/(d^2*abs(c))

________________________________________________________________________________________

Mupad [B]
time = 1.04, size = 110, normalized size = 0.90 \begin {gather*} -\frac {2\,\left (48\,a^2\,c^2+24\,a\,b^2\,c+192\,a\,b\,c^2\,x+192\,a\,c^3\,x^2-b^4+16\,b^3\,c\,x+144\,b^2\,c^2\,x^2+256\,b\,c^3\,x^3+128\,c^4\,x^4\right )}{3\,d^2\,{\left (4\,a\,c-b^2\right )}^3\,\left (b+2\,c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(48*a^2*c^2 - b^4 + 128*c^4*x^4 + 192*a*c^3*x^2 + 256*b*c^3*x^3 + 144*b^2*c^2*x^2 + 24*a*b^2*c + 16*b^3*c*
x + 192*a*b*c^2*x))/(3*d^2*(4*a*c - b^2)^3*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))

________________________________________________________________________________________