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

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

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Rubi [A]  time = 0.17, 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 = {78, 51, 63, 208} \begin {gather*} -\frac {2 d^2 (b c-a d)}{\sqrt {e+f x} (d e-c f)^4}+\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 (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 51

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 63

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 78

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] || LtQ
[p, n]))))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, 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 ) \operatorname {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 [C]  time = 0.06, size = 86, normalized size = 0.46 \begin {gather*} -\frac {2 \left (7 f (e+f x) (b c-a d) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\frac {d (e+f x)}{d e-c f}\right )+5 (b e-a f) (d e-c f)\right )}{35 f (e+f x)^{7/2} (d e-c f)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*(b*e - a*f)*(d*e - c*f) + 7*(b*c - a*d)*f*(e + f*x)*Hypergeometric2F1[-5/2, 1, -3/2, (d*(e + f*x))/(d*e
- c*f)]))/(35*f*(d*e - c*f)^2*(e + f*x)^(7/2))

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IntegrateAlgebraic [A]  time = 0.40, size = 369, normalized size = 1.99 \begin {gather*} -\frac {2 \left (15 a c^3 f^4-21 a c^2 d f^3 (e+f x)-45 a c^2 d e f^3+45 a c d^2 e^2 f^2+42 a c d^2 e f^2 (e+f x)+35 a c d^2 f^2 (e+f x)^2-15 a d^3 e^3 f-21 a d^3 e^2 f (e+f x)-35 a d^3 e f (e+f x)^2-105 a d^3 f (e+f x)^3+21 b c^3 f^3 (e+f x)-15 b c^3 e f^3+45 b c^2 d e^2 f^2-42 b c^2 d e f^2 (e+f x)-35 b c^2 d f^2 (e+f x)^2-45 b c d^2 e^3 f+21 b c d^2 e^2 f (e+f x)+35 b c d^2 e f (e+f x)^2+105 b c d^2 f (e+f x)^3+15 b d^3 e^4\right )}{105 f (e+f x)^{7/2} (c f-d e)^4}-\frac {2 \left (a d^{7/2}-b c d^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{(c f-d e)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*b*d^3*e^4 - 45*b*c*d^2*e^3*f - 15*a*d^3*e^3*f + 45*b*c^2*d*e^2*f^2 + 45*a*c*d^2*e^2*f^2 - 15*b*c^3*e*f
^3 - 45*a*c^2*d*e*f^3 + 15*a*c^3*f^4 + 21*b*c*d^2*e^2*f*(e + f*x) - 21*a*d^3*e^2*f*(e + f*x) - 42*b*c^2*d*e*f^
2*(e + f*x) + 42*a*c*d^2*e*f^2*(e + f*x) + 21*b*c^3*f^3*(e + f*x) - 21*a*c^2*d*f^3*(e + f*x) + 35*b*c*d^2*e*f*
(e + f*x)^2 - 35*a*d^3*e*f*(e + f*x)^2 - 35*b*c^2*d*f^2*(e + f*x)^2 + 35*a*c*d^2*f^2*(e + f*x)^2 + 105*b*c*d^2
*f*(e + f*x)^3 - 105*a*d^3*f*(e + f*x)^3))/(105*f*(-(d*e) + c*f)^4*(e + f*x)^(7/2)) - (2*(-(b*c*d^(5/2)) + a*d
^(7/2))*ArcTan[(Sqrt[d]*Sqrt[-(d*e) + c*f]*Sqrt[e + f*x])/(d*e - c*f)])/(-(d*e) + c*f)^(9/2)

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fricas [B]  time = 1.29, size = 1396, normalized size = 7.55

result too large to display

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)*e*f^4*x^3 + 6*(b*c*d^2 - a*d^3)*e^2*f^3*x^2 + 4*
(b*c*d^2 - a*d^3)*e^3*f^2*x + (b*c*d^2 - a*d^3)*e^4*f)*sqrt(d/(d*e - c*f))*log((d*f*x + 2*d*e - c*f - 2*(d*e -
c*f)*sqrt(f*x + e)*sqrt(d/(d*e - c*f)))/(d*x + c)) + 2*(15*b*d^3*e^4 + 15*a*c^3*f^4 + 105*(b*c*d^2 - a*d^3)*f
^4*x^3 + 4*(29*b*c*d^2 - 44*a*d^3)*e^3*f - 2*(16*b*c^2*d - 61*a*c*d^2)*e^2*f^2 + 6*(b*c^3 - 11*a*c^2*d)*e*f^3
+ 35*(10*(b*c*d^2 - a*d^3)*e*f^3 - (b*c^2*d - a*c*d^2)*f^4)*x^2 + 7*(58*(b*c*d^2 - a*d^3)*e^2*f^2 - 16*(b*c^2*
d - a*c*d^2)*e*f^3 + 3*(b*c^3 - a*c^2*d)*f^4)*x)*sqrt(f*x + e))/(d^4*e^8*f - 4*c*d^3*e^7*f^2 + 6*c^2*d^2*e^6*f
^3 - 4*c^3*d*e^5*f^4 + c^4*e^4*f^5 + (d^4*e^4*f^5 - 4*c*d^3*e^3*f^6 + 6*c^2*d^2*e^2*f^7 - 4*c^3*d*e*f^8 + c^4*
f^9)*x^4 + 4*(d^4*e^5*f^4 - 4*c*d^3*e^4*f^5 + 6*c^2*d^2*e^3*f^6 - 4*c^3*d*e^2*f^7 + c^4*e*f^8)*x^3 + 6*(d^4*e^
6*f^3 - 4*c*d^3*e^5*f^4 + 6*c^2*d^2*e^4*f^5 - 4*c^3*d*e^3*f^6 + c^4*e^2*f^7)*x^2 + 4*(d^4*e^7*f^2 - 4*c*d^3*e^
6*f^3 + 6*c^2*d^2*e^5*f^4 - 4*c^3*d*e^4*f^5 + c^4*e^3*f^6)*x), 2/105*(105*((b*c*d^2 - a*d^3)*f^5*x^4 + 4*(b*c*
d^2 - a*d^3)*e*f^4*x^3 + 6*(b*c*d^2 - a*d^3)*e^2*f^3*x^2 + 4*(b*c*d^2 - a*d^3)*e^3*f^2*x + (b*c*d^2 - a*d^3)*e
^4*f)*sqrt(-d/(d*e - c*f))*arctan(-(d*e - c*f)*sqrt(f*x + e)*sqrt(-d/(d*e - c*f))/(d*f*x + d*e)) - (15*b*d^3*e
^4 + 15*a*c^3*f^4 + 105*(b*c*d^2 - a*d^3)*f^4*x^3 + 4*(29*b*c*d^2 - 44*a*d^3)*e^3*f - 2*(16*b*c^2*d - 61*a*c*d
^2)*e^2*f^2 + 6*(b*c^3 - 11*a*c^2*d)*e*f^3 + 35*(10*(b*c*d^2 - a*d^3)*e*f^3 - (b*c^2*d - a*c*d^2)*f^4)*x^2 + 7
*(58*(b*c*d^2 - a*d^3)*e^2*f^2 - 16*(b*c^2*d - a*c*d^2)*e*f^3 + 3*(b*c^3 - a*c^2*d)*f^4)*x)*sqrt(f*x + e))/(d^
4*e^8*f - 4*c*d^3*e^7*f^2 + 6*c^2*d^2*e^6*f^3 - 4*c^3*d*e^5*f^4 + c^4*e^4*f^5 + (d^4*e^4*f^5 - 4*c*d^3*e^3*f^6
+ 6*c^2*d^2*e^2*f^7 - 4*c^3*d*e*f^8 + c^4*f^9)*x^4 + 4*(d^4*e^5*f^4 - 4*c*d^3*e^4*f^5 + 6*c^2*d^2*e^3*f^6 - 4
*c^3*d*e^2*f^7 + c^4*e*f^8)*x^3 + 6*(d^4*e^6*f^3 - 4*c*d^3*e^5*f^4 + 6*c^2*d^2*e^4*f^5 - 4*c^3*d*e^3*f^6 + c^4
*e^2*f^7)*x^2 + 4*(d^4*e^7*f^2 - 4*c*d^3*e^6*f^3 + 6*c^2*d^2*e^5*f^4 - 4*c^3*d*e^4*f^5 + c^4*e^3*f^6)*x)]

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giac [B]  time = 1.29, 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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maple [A]  time = 0.02, size = 281, normalized size = 1.52 \begin {gather*} \frac {2 a \,d^{4} \arctan \left (\frac {\sqrt {f x +e}\, d}{\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 b c \,d^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\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 a \,d^{3}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}-\frac {2 b c \,d^{2}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}-\frac {2 a \,d^{2}}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 b c d}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 a d}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 b c}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 a}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}+\frac {2 b e}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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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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sympy [A]  time = 60.09, 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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