3.15.9 \(\int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx\) [1409]

Optimal. Leaf size=162 \[ -\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac {5 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}} \]

[Out]

-5/24*d*(d*x+c)^(3/2)/b^2/(b*x+a)^3-1/4*(d*x+c)^(5/2)/b/(b*x+a)^4+5/64*d^4*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d
+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(3/2)-5/32*d^2*(d*x+c)^(1/2)/b^3/(b*x+a)^2-5/64*d^3*(d*x+c)^(1/2)/b^3/(-a*d+b*
c)/(b*x+a)

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Rubi [A]
time = 0.05, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {43, 44, 65, 214} \begin {gather*} \frac {5 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}}-\frac {5 d^3 \sqrt {c+d x}}{64 b^3 (a+b x) (b c-a d)}-\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^5,x]

[Out]

(-5*d^2*Sqrt[c + d*x])/(32*b^3*(a + b*x)^2) - (5*d^3*Sqrt[c + d*x])/(64*b^3*(b*c - a*d)*(a + b*x)) - (5*d*(c +
d*x)^(3/2))/(24*b^2*(a + b*x)^3) - (c + d*x)^(5/2)/(4*b*(a + b*x)^4) + (5*d^4*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])
/Sqrt[b*c - a*d]])/(64*b^(7/2)*(b*c - a*d)^(3/2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/2}}{(a+b x)^5} \, dx &=-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac {(5 d) \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac {\left (5 d^2\right ) \int \frac {\sqrt {c+d x}}{(a+b x)^3} \, dx}{16 b^2}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac {\left (5 d^3\right ) \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{64 b^3}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}-\frac {\left (5 d^4\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{128 b^3 (b c-a d)}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}-\frac {\left (5 d^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{64 b^3 (b c-a d)}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{32 b^3 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac {5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac {5 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.97, size = 172, normalized size = 1.06 \begin {gather*} \frac {\sqrt {c+d x} \left (15 a^3 d^3+5 a^2 b d^2 (2 c+11 d x)+a b^2 d \left (8 c^2+36 c d x+73 d^2 x^2\right )-b^3 \left (48 c^3+136 c^2 d x+118 c d^2 x^2+15 d^3 x^3\right )\right )}{192 b^3 (b c-a d) (a+b x)^4}+\frac {5 d^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{64 b^{7/2} (-b c+a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(15*a^3*d^3 + 5*a^2*b*d^2*(2*c + 11*d*x) + a*b^2*d*(8*c^2 + 36*c*d*x + 73*d^2*x^2) - b^3*(48*c^
3 + 136*c^2*d*x + 118*c*d^2*x^2 + 15*d^3*x^3)))/(192*b^3*(b*c - a*d)*(a + b*x)^4) + (5*d^4*ArcTan[(Sqrt[b]*Sqr
t[c + d*x])/Sqrt[-(b*c) + a*d]])/(64*b^(7/2)*(-(b*c) + a*d)^(3/2))

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Maple [A]
time = 0.16, size = 159, normalized size = 0.98

method	result	size
derivativedivides	\(2 d^{4} \left (\frac {\frac {5 \left (d x +c \right )^{\frac {7}{2}}}{128 \left (a d -b c \right )}-\frac {73 \left (d x +c \right )^{\frac {5}{2}}}{384 b}-\frac {55 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{384 b^{2}}-\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d x +c}}{128 b^{3}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{4}}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a d -b c \right ) b^{3} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(159\)
default	\(2 d^{4} \left (\frac {\frac {5 \left (d x +c \right )^{\frac {7}{2}}}{128 \left (a d -b c \right )}-\frac {73 \left (d x +c \right )^{\frac {5}{2}}}{384 b}-\frac {55 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{384 b^{2}}-\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d x +c}}{128 b^{3}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{4}}+\frac {5 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a d -b c \right ) b^{3} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^4*((5/128/(a*d-b*c)*(d*x+c)^(7/2)-73/384*(d*x+c)^(5/2)/b-55/384*(a*d-b*c)/b^2*(d*x+c)^(3/2)-5/128/b^3*(a^2
*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)^4+5/128/(a*d-b*c)/b^3/((a*d-b*c)*b)^(1/2)*arctan(b*
(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))

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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((d*x+c)^(5/2)/(b*x+a)^5,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(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (134) = 268\).
time = 0.42, size = 894, normalized size = 5.52 \begin {gather*} \left [-\frac {15 \, {\left (b^{4} d^{4} x^{4} + 4 \, a b^{3} d^{4} x^{3} + 6 \, a^{2} b^{2} d^{4} x^{2} + 4 \, a^{3} b d^{4} x + a^{4} d^{4}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (48 \, b^{5} c^{4} - 56 \, a b^{4} c^{3} d - 2 \, a^{2} b^{3} c^{2} d^{2} - 5 \, a^{3} b^{2} c d^{3} + 15 \, a^{4} b d^{4} + 15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )} x^{3} + {\left (118 \, b^{5} c^{2} d^{2} - 191 \, a b^{4} c d^{3} + 73 \, a^{2} b^{3} d^{4}\right )} x^{2} + {\left (136 \, b^{5} c^{3} d - 172 \, a b^{4} c^{2} d^{2} - 19 \, a^{2} b^{3} c d^{3} + 55 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{384 \, {\left (a^{4} b^{6} c^{2} - 2 \, a^{5} b^{5} c d + a^{6} b^{4} d^{2} + {\left (b^{10} c^{2} - 2 \, a b^{9} c d + a^{2} b^{8} d^{2}\right )} x^{4} + 4 \, {\left (a b^{9} c^{2} - 2 \, a^{2} b^{8} c d + a^{3} b^{7} d^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} c^{2} - 2 \, a^{3} b^{7} c d + a^{4} b^{6} d^{2}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} c^{2} - 2 \, a^{4} b^{6} c d + a^{5} b^{5} d^{2}\right )} x\right )}}, -\frac {15 \, {\left (b^{4} d^{4} x^{4} + 4 \, a b^{3} d^{4} x^{3} + 6 \, a^{2} b^{2} d^{4} x^{2} + 4 \, a^{3} b d^{4} x + a^{4} d^{4}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) + {\left (48 \, b^{5} c^{4} - 56 \, a b^{4} c^{3} d - 2 \, a^{2} b^{3} c^{2} d^{2} - 5 \, a^{3} b^{2} c d^{3} + 15 \, a^{4} b d^{4} + 15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )} x^{3} + {\left (118 \, b^{5} c^{2} d^{2} - 191 \, a b^{4} c d^{3} + 73 \, a^{2} b^{3} d^{4}\right )} x^{2} + {\left (136 \, b^{5} c^{3} d - 172 \, a b^{4} c^{2} d^{2} - 19 \, a^{2} b^{3} c d^{3} + 55 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{192 \, {\left (a^{4} b^{6} c^{2} - 2 \, a^{5} b^{5} c d + a^{6} b^{4} d^{2} + {\left (b^{10} c^{2} - 2 \, a b^{9} c d + a^{2} b^{8} d^{2}\right )} x^{4} + 4 \, {\left (a b^{9} c^{2} - 2 \, a^{2} b^{8} c d + a^{3} b^{7} d^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} c^{2} - 2 \, a^{3} b^{7} c d + a^{4} b^{6} d^{2}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} c^{2} - 2 \, a^{4} b^{6} c d + a^{5} b^{5} d^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(b^2*c - a*b*d)*
log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(48*b^5*c^4 - 56*a*b^4*c^3*d -
2*a^2*b^3*c^2*d^2 - 5*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (118*b^5*c^2*d^2 - 191*a
*b^4*c*d^3 + 73*a^2*b^3*d^4)*x^2 + (136*b^5*c^3*d - 172*a*b^4*c^2*d^2 - 19*a^2*b^3*c*d^3 + 55*a^3*b^2*d^4)*x)*
sqrt(d*x + c))/(a^4*b^6*c^2 - 2*a^5*b^5*c*d + a^6*b^4*d^2 + (b^10*c^2 - 2*a*b^9*c*d + a^2*b^8*d^2)*x^4 + 4*(a*
b^9*c^2 - 2*a^2*b^8*c*d + a^3*b^7*d^2)*x^3 + 6*(a^2*b^8*c^2 - 2*a^3*b^7*c*d + a^4*b^6*d^2)*x^2 + 4*(a^3*b^7*c^
2 - 2*a^4*b^6*c*d + a^5*b^5*d^2)*x), -1/192*(15*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d
^4*x + a^4*d^4)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (48*b^5*c^4 -
56*a*b^4*c^3*d - 2*a^2*b^3*c^2*d^2 - 5*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (118*b^
5*c^2*d^2 - 191*a*b^4*c*d^3 + 73*a^2*b^3*d^4)*x^2 + (136*b^5*c^3*d - 172*a*b^4*c^2*d^2 - 19*a^2*b^3*c*d^3 + 55
*a^3*b^2*d^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^2 - 2*a^5*b^5*c*d + a^6*b^4*d^2 + (b^10*c^2 - 2*a*b^9*c*d + a^2*b^8
*d^2)*x^4 + 4*(a*b^9*c^2 - 2*a^2*b^8*c*d + a^3*b^7*d^2)*x^3 + 6*(a^2*b^8*c^2 - 2*a^3*b^7*c*d + a^4*b^6*d^2)*x^
2 + 4*(a^3*b^7*c^2 - 2*a^4*b^6*c*d + a^5*b^5*d^2)*x)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 2.53, size = 259, normalized size = 1.60 \begin {gather*} -\frac {5 \, d^{4} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{64 \, {\left (b^{4} c - a b^{3} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {15 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d^{4} + 73 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} c d^{4} - 55 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} + 15 \, \sqrt {d x + c} b^{3} c^{3} d^{4} - 73 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{2} d^{5} + 110 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} c d^{5} - 45 \, \sqrt {d x + c} a b^{2} c^{2} d^{5} - 55 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b d^{6} + 45 \, \sqrt {d x + c} a^{2} b c d^{6} - 15 \, \sqrt {d x + c} a^{3} d^{7}}{192 \, {\left (b^{4} c - a b^{3} d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c - a*b^3*d)*sqrt(-b^2*c + a*b*d)) - 1/192*(15*(d
*x + c)^(7/2)*b^3*d^4 + 73*(d*x + c)^(5/2)*b^3*c*d^4 - 55*(d*x + c)^(3/2)*b^3*c^2*d^4 + 15*sqrt(d*x + c)*b^3*c
^3*d^4 - 73*(d*x + c)^(5/2)*a*b^2*d^5 + 110*(d*x + c)^(3/2)*a*b^2*c*d^5 - 45*sqrt(d*x + c)*a*b^2*c^2*d^5 - 55*
(d*x + c)^(3/2)*a^2*b*d^6 + 45*sqrt(d*x + c)*a^2*b*c*d^6 - 15*sqrt(d*x + c)*a^3*d^7)/((b^4*c - a*b^3*d)*((d*x
+ c)*b - b*c + a*d)^4)

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Mupad [B]
time = 0.41, size = 309, normalized size = 1.91 \begin {gather*} \frac {5\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,b^{7/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {\frac {73\,d^4\,{\left (c+d\,x\right )}^{5/2}}{192\,b}-\frac {5\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,\left (a\,d-b\,c\right )}+\frac {5\,d^4\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{64\,b^3}+\frac {55\,d^4\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{192\,b^2}}{b^4\,{\left (c+d\,x\right )}^4-\left (4\,b^4\,c-4\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^3-\left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )+a^4\,d^4+b^4\,c^4+{\left (c+d\,x\right )}^2\,\left (6\,a^2\,b^2\,d^2-12\,a\,b^3\,c\,d+6\,b^4\,c^2\right )+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d-4\,a^3\,b\,c\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x)^5,x)

[Out]

(5*d^4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(7/2)*(a*d - b*c)^(3/2)) - ((73*d^4*(c + d*x)^
(5/2))/(192*b) - (5*d^4*(c + d*x)^(7/2))/(64*(a*d - b*c)) + (5*d^4*(c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*
c*d))/(64*b^3) + (55*d^4*(a*d - b*c)*(c + d*x)^(3/2))/(192*b^2))/(b^4*(c + d*x)^4 - (4*b^4*c - 4*a*b^3*d)*(c +
d*x)^3 - (c + d*x)*(4*b^4*c^3 - 4*a^3*b*d^3 + 12*a^2*b^2*c*d^2 - 12*a*b^3*c^2*d) + a^4*d^4 + b^4*c^4 + (c + d
*x)^2*(6*b^4*c^2 + 6*a^2*b^2*d^2 - 12*a*b^3*c*d) + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3)

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