3.18.73 \(\int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx\) [1773]

Optimal. Leaf size=185 \[ -\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {2 d (b c-a d)}{3 (d e-c f)^3 (e+f x)^{3/2}}-\frac {2 d^2 (b c-a d)}{(d e-c f)^4 \sqrt {e+f x}}+\frac {2 d^{5/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{9/2}} \]

[Out]

-2/7*(-a*f+b*e)/f/(-c*f+d*e)/(f*x+e)^(7/2)-2/5*(-a*d+b*c)/(-c*f+d*e)^2/(f*x+e)^(5/2)-2/3*d*(-a*d+b*c)/(-c*f+d*
e)^3/(f*x+e)^(3/2)+2*d^(5/2)*(-a*d+b*c)*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/(-c*f+d*e)^(9/2)-2*d^2
*(-a*d+b*c)/(-c*f+d*e)^4/(f*x+e)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {79, 53, 65, 214} \begin {gather*} \frac {2 d^{5/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{9/2}}-\frac {2 d^2 (b c-a d)}{\sqrt {e+f x} (d e-c f)^4}-\frac {2 d (b c-a d)}{3 (e+f x)^{3/2} (d e-c f)^3}-\frac {2 (b c-a d)}{5 (e+f x)^{5/2} (d e-c f)^2}-\frac {2 (b e-a f)}{7 f (e+f x)^{7/2} (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((c + d*x)*(e + f*x)^(9/2)),x]

[Out]

(-2*(b*e - a*f))/(7*f*(d*e - c*f)*(e + f*x)^(7/2)) - (2*(b*c - a*d))/(5*(d*e - c*f)^2*(e + f*x)^(5/2)) - (2*d*
(b*c - a*d))/(3*(d*e - c*f)^3*(e + f*x)^(3/2)) - (2*d^2*(b*c - a*d))/((d*e - c*f)^4*Sqrt[e + f*x]) + (2*d^(5/2
)*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)^(9/2)

Rule 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx &=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {(b c-a d) \int \frac {1}{(c+d x) (e+f x)^{7/2}} \, dx}{d e-c f}\\ &=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {(d (b c-a d)) \int \frac {1}{(c+d x) (e+f x)^{5/2}} \, dx}{(d e-c f)^2}\\ &=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {2 d (b c-a d)}{3 (d e-c f)^3 (e+f x)^{3/2}}-\frac {\left (d^2 (b c-a d)\right ) \int \frac {1}{(c+d x) (e+f x)^{3/2}} \, dx}{(d e-c f)^3}\\ &=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {2 d (b c-a d)}{3 (d e-c f)^3 (e+f x)^{3/2}}-\frac {2 d^2 (b c-a d)}{(d e-c f)^4 \sqrt {e+f x}}-\frac {\left (d^3 (b c-a d)\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{(d e-c f)^4}\\ &=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {2 d (b c-a d)}{3 (d e-c f)^3 (e+f x)^{3/2}}-\frac {2 d^2 (b c-a d)}{(d e-c f)^4 \sqrt {e+f x}}-\frac {\left (2 d^3 (b c-a d)\right ) \text {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{f (d e-c f)^4}\\ &=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {2 d (b c-a d)}{3 (d e-c f)^3 (e+f x)^{3/2}}-\frac {2 d^2 (b c-a d)}{(d e-c f)^4 \sqrt {e+f x}}+\frac {2 d^{5/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 265, normalized size = 1.43 \begin {gather*} \frac {-2 b \left (15 d^3 e^4+3 c^3 f^3 (2 e+7 f x)-c^2 d f^2 \left (32 e^2+112 e f x+35 f^2 x^2\right )+c d^2 f \left (116 e^3+406 e^2 f x+350 e f^2 x^2+105 f^3 x^3\right )\right )+2 a f \left (-15 c^3 f^3+3 c^2 d f^2 (22 e+7 f x)-c d^2 f \left (122 e^2+112 e f x+35 f^2 x^2\right )+d^3 \left (176 e^3+406 e^2 f x+350 e f^2 x^2+105 f^3 x^3\right )\right )}{105 f (d e-c f)^4 (e+f x)^{7/2}}+\frac {2 d^{5/2} (-b c+a d) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((c + d*x)*(e + f*x)^(9/2)),x]

[Out]

(-2*b*(15*d^3*e^4 + 3*c^3*f^3*(2*e + 7*f*x) - c^2*d*f^2*(32*e^2 + 112*e*f*x + 35*f^2*x^2) + c*d^2*f*(116*e^3 +
406*e^2*f*x + 350*e*f^2*x^2 + 105*f^3*x^3)) + 2*a*f*(-15*c^3*f^3 + 3*c^2*d*f^2*(22*e + 7*f*x) - c*d^2*f*(122*
e^2 + 112*e*f*x + 35*f^2*x^2) + d^3*(176*e^3 + 406*e^2*f*x + 350*e*f^2*x^2 + 105*f^3*x^3)))/(105*f*(d*e - c*f)
^4*(e + f*x)^(7/2)) + (2*d^(5/2)*(-(b*c) + a*d)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(-(d*e) +
c*f)^(9/2)

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Maple [A]
time = 0.10, size = 178, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 d^{3} f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a f -b e \right )}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}-\frac {2 f \left (a d -b c \right ) d}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 f \left (a d -b c \right )}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}+\frac {2 f \left (a d -b c \right ) d^{2}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}}{f}\)	\(178\)
default	\(\frac {\frac {2 d^{3} f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a f -b e \right )}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}-\frac {2 f \left (a d -b c \right ) d}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 f \left (a d -b c \right )}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}+\frac {2 f \left (a d -b c \right ) d^{2}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}}{f}\)	\(178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/f*(d^3*f*(a*d-b*c)/(c*f-d*e)^4/((c*f-d*e)*d)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2))-1/7*(a*f-b*e)
/(c*f-d*e)/(f*x+e)^(7/2)-1/3*f*(a*d-b*c)/(c*f-d*e)^3*d/(f*x+e)^(3/2)+1/5*f*(a*d-b*c)/(c*f-d*e)^2/(f*x+e)^(5/2)
+f*(a*d-b*c)/(c*f-d*e)^4*d^2/(f*x+e)^(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((b*x+a)/(d*x+c)/(f*x+e)^(9/2),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(c*f-%e*d>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 691 vs. \(2 (173) = 346\).
time = 1.18, size = 1397, normalized size = 7.55 \begin {gather*} \left [-\frac {105 \, {\left ({\left (b c d^{2} - a d^{3}\right )} f^{5} x^{4} + 4 \, {\left (b c d^{2} - a d^{3}\right )} f^{4} x^{3} e + 6 \, {\left (b c d^{2} - a d^{3}\right )} f^{3} x^{2} e^{2} + 4 \, {\left (b c d^{2} - a d^{3}\right )} f^{2} x e^{3} + {\left (b c d^{2} - a d^{3}\right )} f e^{4}\right )} \sqrt {-\frac {d}{c f - d e}} \log \left (\frac {d f x - c f + 2 \, {\left (c f - d e\right )} \sqrt {f x + e} \sqrt {-\frac {d}{c f - d e}} + 2 \, d e}{d x + c}\right ) + 2 \, {\left (15 \, a c^{3} f^{4} + 105 \, {\left (b c d^{2} - a d^{3}\right )} f^{4} x^{3} - 35 \, {\left (b c^{2} d - a c d^{2}\right )} f^{4} x^{2} + 21 \, {\left (b c^{3} - a c^{2} d\right )} f^{4} x + 15 \, b d^{3} e^{4} + 4 \, {\left (29 \, b c d^{2} - 44 \, a d^{3}\right )} f e^{3} + 2 \, {\left (203 \, {\left (b c d^{2} - a d^{3}\right )} f^{2} x - {\left (16 \, b c^{2} d - 61 \, a c d^{2}\right )} f^{2}\right )} e^{2} + 2 \, {\left (175 \, {\left (b c d^{2} - a d^{3}\right )} f^{3} x^{2} - 56 \, {\left (b c^{2} d - a c d^{2}\right )} f^{3} x + 3 \, {\left (b c^{3} - 11 \, a c^{2} d\right )} f^{3}\right )} e\right )} \sqrt {f x + e}}{105 \, {\left (c^{4} f^{9} x^{4} + d^{4} f e^{8} + 4 \, {\left (d^{4} f^{2} x - c d^{3} f^{2}\right )} e^{7} + 2 \, {\left (3 \, d^{4} f^{3} x^{2} - 8 \, c d^{3} f^{3} x + 3 \, c^{2} d^{2} f^{3}\right )} e^{6} + 4 \, {\left (d^{4} f^{4} x^{3} - 6 \, c d^{3} f^{4} x^{2} + 6 \, c^{2} d^{2} f^{4} x - c^{3} d f^{4}\right )} e^{5} + {\left (d^{4} f^{5} x^{4} - 16 \, c d^{3} f^{5} x^{3} + 36 \, c^{2} d^{2} f^{5} x^{2} - 16 \, c^{3} d f^{5} x + c^{4} f^{5}\right )} e^{4} - 4 \, {\left (c d^{3} f^{6} x^{4} - 6 \, c^{2} d^{2} f^{6} x^{3} + 6 \, c^{3} d f^{6} x^{2} - c^{4} f^{6} x\right )} e^{3} + 2 \, {\left (3 \, c^{2} d^{2} f^{7} x^{4} - 8 \, c^{3} d f^{7} x^{3} + 3 \, c^{4} f^{7} x^{2}\right )} e^{2} - 4 \, {\left (c^{3} d f^{8} x^{4} - c^{4} f^{8} x^{3}\right )} e\right )}}, -\frac {2 \, {\left (105 \, {\left ({\left (b c d^{2} - a d^{3}\right )} f^{5} x^{4} + 4 \, {\left (b c d^{2} - a d^{3}\right )} f^{4} x^{3} e + 6 \, {\left (b c d^{2} - a d^{3}\right )} f^{3} x^{2} e^{2} + 4 \, {\left (b c d^{2} - a d^{3}\right )} f^{2} x e^{3} + {\left (b c d^{2} - a d^{3}\right )} f e^{4}\right )} \sqrt {\frac {d}{c f - d e}} \arctan \left (-\frac {{\left (c f - d e\right )} \sqrt {f x + e} \sqrt {\frac {d}{c f - d e}}}{d f x + d e}\right ) + {\left (15 \, a c^{3} f^{4} + 105 \, {\left (b c d^{2} - a d^{3}\right )} f^{4} x^{3} - 35 \, {\left (b c^{2} d - a c d^{2}\right )} f^{4} x^{2} + 21 \, {\left (b c^{3} - a c^{2} d\right )} f^{4} x + 15 \, b d^{3} e^{4} + 4 \, {\left (29 \, b c d^{2} - 44 \, a d^{3}\right )} f e^{3} + 2 \, {\left (203 \, {\left (b c d^{2} - a d^{3}\right )} f^{2} x - {\left (16 \, b c^{2} d - 61 \, a c d^{2}\right )} f^{2}\right )} e^{2} + 2 \, {\left (175 \, {\left (b c d^{2} - a d^{3}\right )} f^{3} x^{2} - 56 \, {\left (b c^{2} d - a c d^{2}\right )} f^{3} x + 3 \, {\left (b c^{3} - 11 \, a c^{2} d\right )} f^{3}\right )} e\right )} \sqrt {f x + e}\right )}}{105 \, {\left (c^{4} f^{9} x^{4} + d^{4} f e^{8} + 4 \, {\left (d^{4} f^{2} x - c d^{3} f^{2}\right )} e^{7} + 2 \, {\left (3 \, d^{4} f^{3} x^{2} - 8 \, c d^{3} f^{3} x + 3 \, c^{2} d^{2} f^{3}\right )} e^{6} + 4 \, {\left (d^{4} f^{4} x^{3} - 6 \, c d^{3} f^{4} x^{2} + 6 \, c^{2} d^{2} f^{4} x - c^{3} d f^{4}\right )} e^{5} + {\left (d^{4} f^{5} x^{4} - 16 \, c d^{3} f^{5} x^{3} + 36 \, c^{2} d^{2} f^{5} x^{2} - 16 \, c^{3} d f^{5} x + c^{4} f^{5}\right )} e^{4} - 4 \, {\left (c d^{3} f^{6} x^{4} - 6 \, c^{2} d^{2} f^{6} x^{3} + 6 \, c^{3} d f^{6} x^{2} - c^{4} f^{6} x\right )} e^{3} + 2 \, {\left (3 \, c^{2} d^{2} f^{7} x^{4} - 8 \, c^{3} d f^{7} x^{3} + 3 \, c^{4} f^{7} x^{2}\right )} e^{2} - 4 \, {\left (c^{3} d f^{8} x^{4} - c^{4} f^{8} x^{3}\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x, algorithm="fricas")

[Out]

[-1/105*(105*((b*c*d^2 - a*d^3)*f^5*x^4 + 4*(b*c*d^2 - a*d^3)*f^4*x^3*e + 6*(b*c*d^2 - a*d^3)*f^3*x^2*e^2 + 4*
(b*c*d^2 - a*d^3)*f^2*x*e^3 + (b*c*d^2 - a*d^3)*f*e^4)*sqrt(-d/(c*f - d*e))*log((d*f*x - c*f + 2*(c*f - d*e)*s
qrt(f*x + e)*sqrt(-d/(c*f - d*e)) + 2*d*e)/(d*x + c)) + 2*(15*a*c^3*f^4 + 105*(b*c*d^2 - a*d^3)*f^4*x^3 - 35*(
b*c^2*d - a*c*d^2)*f^4*x^2 + 21*(b*c^3 - a*c^2*d)*f^4*x + 15*b*d^3*e^4 + 4*(29*b*c*d^2 - 44*a*d^3)*f*e^3 + 2*(
203*(b*c*d^2 - a*d^3)*f^2*x - (16*b*c^2*d - 61*a*c*d^2)*f^2)*e^2 + 2*(175*(b*c*d^2 - a*d^3)*f^3*x^2 - 56*(b*c^
2*d - a*c*d^2)*f^3*x + 3*(b*c^3 - 11*a*c^2*d)*f^3)*e)*sqrt(f*x + e))/(c^4*f^9*x^4 + d^4*f*e^8 + 4*(d^4*f^2*x -
c*d^3*f^2)*e^7 + 2*(3*d^4*f^3*x^2 - 8*c*d^3*f^3*x + 3*c^2*d^2*f^3)*e^6 + 4*(d^4*f^4*x^3 - 6*c*d^3*f^4*x^2 + 6
*c^2*d^2*f^4*x - c^3*d*f^4)*e^5 + (d^4*f^5*x^4 - 16*c*d^3*f^5*x^3 + 36*c^2*d^2*f^5*x^2 - 16*c^3*d*f^5*x + c^4*
f^5)*e^4 - 4*(c*d^3*f^6*x^4 - 6*c^2*d^2*f^6*x^3 + 6*c^3*d*f^6*x^2 - c^4*f^6*x)*e^3 + 2*(3*c^2*d^2*f^7*x^4 - 8*
c^3*d*f^7*x^3 + 3*c^4*f^7*x^2)*e^2 - 4*(c^3*d*f^8*x^4 - c^4*f^8*x^3)*e), -2/105*(105*((b*c*d^2 - a*d^3)*f^5*x^
4 + 4*(b*c*d^2 - a*d^3)*f^4*x^3*e + 6*(b*c*d^2 - a*d^3)*f^3*x^2*e^2 + 4*(b*c*d^2 - a*d^3)*f^2*x*e^3 + (b*c*d^2
- a*d^3)*f*e^4)*sqrt(d/(c*f - d*e))*arctan(-(c*f - d*e)*sqrt(f*x + e)*sqrt(d/(c*f - d*e))/(d*f*x + d*e)) + (1
5*a*c^3*f^4 + 105*(b*c*d^2 - a*d^3)*f^4*x^3 - 35*(b*c^2*d - a*c*d^2)*f^4*x^2 + 21*(b*c^3 - a*c^2*d)*f^4*x + 15
*b*d^3*e^4 + 4*(29*b*c*d^2 - 44*a*d^3)*f*e^3 + 2*(203*(b*c*d^2 - a*d^3)*f^2*x - (16*b*c^2*d - 61*a*c*d^2)*f^2)
*e^2 + 2*(175*(b*c*d^2 - a*d^3)*f^3*x^2 - 56*(b*c^2*d - a*c*d^2)*f^3*x + 3*(b*c^3 - 11*a*c^2*d)*f^3)*e)*sqrt(f
*x + e))/(c^4*f^9*x^4 + d^4*f*e^8 + 4*(d^4*f^2*x - c*d^3*f^2)*e^7 + 2*(3*d^4*f^3*x^2 - 8*c*d^3*f^3*x + 3*c^2*d
^2*f^3)*e^6 + 4*(d^4*f^4*x^3 - 6*c*d^3*f^4*x^2 + 6*c^2*d^2*f^4*x - c^3*d*f^4)*e^5 + (d^4*f^5*x^4 - 16*c*d^3*f^
5*x^3 + 36*c^2*d^2*f^5*x^2 - 16*c^3*d*f^5*x + c^4*f^5)*e^4 - 4*(c*d^3*f^6*x^4 - 6*c^2*d^2*f^6*x^3 + 6*c^3*d*f^
6*x^2 - c^4*f^6*x)*e^3 + 2*(3*c^2*d^2*f^7*x^4 - 8*c^3*d*f^7*x^3 + 3*c^4*f^7*x^2)*e^2 - 4*(c^3*d*f^8*x^4 - c^4*
f^8*x^3)*e)]

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Sympy [A]
time = 39.57, size = 168, normalized size = 0.91 \begin {gather*} \frac {2 d^{2} \left (a d - b c\right )}{\sqrt {e + f x} \left (c f - d e\right )^{4}} + \frac {2 d^{2} \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{\sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{4}} - \frac {2 d \left (a d - b c\right )}{3 \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{3}} + \frac {2 \left (a d - b c\right )}{5 \left (e + f x\right )^{\frac {5}{2}} \left (c f - d e\right )^{2}} - \frac {2 \left (a f - b e\right )}{7 f \left (e + f x\right )^{\frac {7}{2}} \left (c f - d e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)**(9/2),x)

[Out]

2*d**2*(a*d - b*c)/(sqrt(e + f*x)*(c*f - d*e)**4) + 2*d**2*(a*d - b*c)*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))
/(sqrt((c*f - d*e)/d)*(c*f - d*e)**4) - 2*d*(a*d - b*c)/(3*(e + f*x)**(3/2)*(c*f - d*e)**3) + 2*(a*d - b*c)/(5
*(e + f*x)**(5/2)*(c*f - d*e)**2) - 2*(a*f - b*e)/(7*f*(e + f*x)**(7/2)*(c*f - d*e))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (173) = 346\).
time = 0.68, size = 450, normalized size = 2.43 \begin {gather*} -\frac {2 \, {\left (b c d^{3} - a d^{4}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c^{4} f^{4} - 4 \, c^{3} d f^{3} e + 6 \, c^{2} d^{2} f^{2} e^{2} - 4 \, c d^{3} f e^{3} + d^{4} e^{4}\right )} \sqrt {c d f - d^{2} e}} - \frac {2 \, {\left (105 \, {\left (f x + e\right )}^{3} b c d^{2} f - 105 \, {\left (f x + e\right )}^{3} a d^{3} f - 35 \, {\left (f x + e\right )}^{2} b c^{2} d f^{2} + 35 \, {\left (f x + e\right )}^{2} a c d^{2} f^{2} + 21 \, {\left (f x + e\right )} b c^{3} f^{3} - 21 \, {\left (f x + e\right )} a c^{2} d f^{3} + 15 \, a c^{3} f^{4} + 35 \, {\left (f x + e\right )}^{2} b c d^{2} f e - 35 \, {\left (f x + e\right )}^{2} a d^{3} f e - 42 \, {\left (f x + e\right )} b c^{2} d f^{2} e + 42 \, {\left (f x + e\right )} a c d^{2} f^{2} e - 15 \, b c^{3} f^{3} e - 45 \, a c^{2} d f^{3} e + 21 \, {\left (f x + e\right )} b c d^{2} f e^{2} - 21 \, {\left (f x + e\right )} a d^{3} f e^{2} + 45 \, b c^{2} d f^{2} e^{2} + 45 \, a c d^{2} f^{2} e^{2} - 45 \, b c d^{2} f e^{3} - 15 \, a d^{3} f e^{3} + 15 \, b d^{3} e^{4}\right )}}{105 \, {\left (c^{4} f^{5} - 4 \, c^{3} d f^{4} e + 6 \, c^{2} d^{2} f^{3} e^{2} - 4 \, c d^{3} f^{2} e^{3} + d^{4} f e^{4}\right )} {\left (f x + e\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x, algorithm="giac")

[Out]

-2*(b*c*d^3 - a*d^4)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^4*f^4 - 4*c^3*d*f^3*e + 6*c^2*d^2*f^2*e^2
- 4*c*d^3*f*e^3 + d^4*e^4)*sqrt(c*d*f - d^2*e)) - 2/105*(105*(f*x + e)^3*b*c*d^2*f - 105*(f*x + e)^3*a*d^3*f
- 35*(f*x + e)^2*b*c^2*d*f^2 + 35*(f*x + e)^2*a*c*d^2*f^2 + 21*(f*x + e)*b*c^3*f^3 - 21*(f*x + e)*a*c^2*d*f^3
+ 15*a*c^3*f^4 + 35*(f*x + e)^2*b*c*d^2*f*e - 35*(f*x + e)^2*a*d^3*f*e - 42*(f*x + e)*b*c^2*d*f^2*e + 42*(f*x
+ e)*a*c*d^2*f^2*e - 15*b*c^3*f^3*e - 45*a*c^2*d*f^3*e + 21*(f*x + e)*b*c*d^2*f*e^2 - 21*(f*x + e)*a*d^3*f*e^2
+ 45*b*c^2*d*f^2*e^2 + 45*a*c*d^2*f^2*e^2 - 45*b*c*d^2*f*e^3 - 15*a*d^3*f*e^3 + 15*b*d^3*e^4)/((c^4*f^5 - 4*c
^3*d*f^4*e + 6*c^2*d^2*f^3*e^2 - 4*c*d^3*f^2*e^3 + d^4*f*e^4)*(f*x + e)^(7/2))

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Mupad [B]
time = 1.36, size = 218, normalized size = 1.18 \begin {gather*} \frac {2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (c^4\,f^4-4\,c^3\,d\,e\,f^3+6\,c^2\,d^2\,e^2\,f^2-4\,c\,d^3\,e^3\,f+d^4\,e^4\right )}{{\left (c\,f-d\,e\right )}^{9/2}}\right )\,\left (a\,d-b\,c\right )}{{\left (c\,f-d\,e\right )}^{9/2}}-\frac {\frac {2\,\left (a\,f-b\,e\right )}{7\,\left (c\,f-d\,e\right )}-\frac {2\,\left (e+f\,x\right )\,\left (a\,d\,f-b\,c\,f\right )}{5\,{\left (c\,f-d\,e\right )}^2}-\frac {2\,d^2\,{\left (e+f\,x\right )}^3\,\left (a\,d\,f-b\,c\,f\right )}{{\left (c\,f-d\,e\right )}^4}+\frac {2\,d\,{\left (e+f\,x\right )}^2\,\left (a\,d\,f-b\,c\,f\right )}{3\,{\left (c\,f-d\,e\right )}^3}}{f\,{\left (e+f\,x\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/((e + f*x)^(9/2)*(c + d*x)),x)

[Out]

(2*d^(5/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(c^4*f^4 + d^4*e^4 + 6*c^2*d^2*e^2*f^2 - 4*c*d^3*e^3*f - 4*c^3*d*e*f^
3))/(c*f - d*e)^(9/2))*(a*d - b*c))/(c*f - d*e)^(9/2) - ((2*(a*f - b*e))/(7*(c*f - d*e)) - (2*(e + f*x)*(a*d*f
- b*c*f))/(5*(c*f - d*e)^2) - (2*d^2*(e + f*x)^3*(a*d*f - b*c*f))/(c*f - d*e)^4 + (2*d*(e + f*x)^2*(a*d*f - b
*c*f))/(3*(c*f - d*e)^3))/(f*(e + f*x)^(7/2))

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