3.1765 \(\int \frac {A+B x}{(a+b x)^3 (d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=281 \[ \frac {7 b^{3/2} e (5 a B e-9 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2}}-\frac {7 b e (5 a B e-9 A b e+4 b B d)}{4 \sqrt {d+e x} (b d-a e)^5}-\frac {7 e (5 a B e-9 A b e+4 b B d)}{12 (d+e x)^{3/2} (b d-a e)^4}-\frac {7 e (5 a B e-9 A b e+4 b B d)}{20 b (d+e x)^{5/2} (b d-a e)^3}-\frac {5 a B e-9 A b e+4 b B d}{4 b (a+b x) (d+e x)^{5/2} (b d-a e)^2}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{5/2} (b d-a e)} \]

[Out]

-7/20*e*(-9*A*b*e+5*B*a*e+4*B*b*d)/b/(-a*e+b*d)^3/(e*x+d)^(5/2)+1/2*(-A*b+B*a)/b/(-a*e+b*d)/(b*x+a)^2/(e*x+d)^
(5/2)+1/4*(9*A*b*e-5*B*a*e-4*B*b*d)/b/(-a*e+b*d)^2/(b*x+a)/(e*x+d)^(5/2)-7/12*e*(-9*A*b*e+5*B*a*e+4*B*b*d)/(-a
*e+b*d)^4/(e*x+d)^(3/2)+7/4*b^(3/2)*e*(-9*A*b*e+5*B*a*e+4*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2
))/(-a*e+b*d)^(11/2)-7/4*b*e*(-9*A*b*e+5*B*a*e+4*B*b*d)/(-a*e+b*d)^5/(e*x+d)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 208} \[ \frac {7 b^{3/2} e (5 a B e-9 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2}}-\frac {7 b e (5 a B e-9 A b e+4 b B d)}{4 \sqrt {d+e x} (b d-a e)^5}-\frac {7 e (5 a B e-9 A b e+4 b B d)}{12 (d+e x)^{3/2} (b d-a e)^4}-\frac {7 e (5 a B e-9 A b e+4 b B d)}{20 b (d+e x)^{5/2} (b d-a e)^3}-\frac {5 a B e-9 A b e+4 b B d}{4 b (a+b x) (d+e x)^{5/2} (b d-a e)^2}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)^3*(d + e*x)^(7/2)),x]

[Out]

(-7*e*(4*b*B*d - 9*A*b*e + 5*a*B*e))/(20*b*(b*d - a*e)^3*(d + e*x)^(5/2)) - (A*b - a*B)/(2*b*(b*d - a*e)*(a +
b*x)^2*(d + e*x)^(5/2)) - (4*b*B*d - 9*A*b*e + 5*a*B*e)/(4*b*(b*d - a*e)^2*(a + b*x)*(d + e*x)^(5/2)) - (7*e*(
4*b*B*d - 9*A*b*e + 5*a*B*e))/(12*(b*d - a*e)^4*(d + e*x)^(3/2)) - (7*b*e*(4*b*B*d - 9*A*b*e + 5*a*B*e))/(4*(b
*d - a*e)^5*Sqrt[d + e*x]) + (7*b^(3/2)*e*(4*b*B*d - 9*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b
*d - a*e]])/(4*(b*d - a*e)^(11/2))

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}{(a+b x)^3 (d+e x)^{7/2}} \, dx &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac {(4 b B d-9 A b e+5 a B e) \int \frac {1}{(a+b x)^2 (d+e x)^{7/2}} \, dx}{4 b (b d-a e)}\\ &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{5/2}}-\frac {(7 e (4 b B d-9 A b e+5 a B e)) \int \frac {1}{(a+b x) (d+e x)^{7/2}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac {7 e (4 b B d-9 A b e+5 a B e)}{20 b (b d-a e)^3 (d+e x)^{5/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{5/2}}-\frac {(7 e (4 b B d-9 A b e+5 a B e)) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^3}\\ &=-\frac {7 e (4 b B d-9 A b e+5 a B e)}{20 b (b d-a e)^3 (d+e x)^{5/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{5/2}}-\frac {7 e (4 b B d-9 A b e+5 a B e)}{12 (b d-a e)^4 (d+e x)^{3/2}}-\frac {(7 b e (4 b B d-9 A b e+5 a B e)) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^4}\\ &=-\frac {7 e (4 b B d-9 A b e+5 a B e)}{20 b (b d-a e)^3 (d+e x)^{5/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{5/2}}-\frac {7 e (4 b B d-9 A b e+5 a B e)}{12 (b d-a e)^4 (d+e x)^{3/2}}-\frac {7 b e (4 b B d-9 A b e+5 a B e)}{4 (b d-a e)^5 \sqrt {d+e x}}-\frac {\left (7 b^2 e (4 b B d-9 A b e+5 a B e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^5}\\ &=-\frac {7 e (4 b B d-9 A b e+5 a B e)}{20 b (b d-a e)^3 (d+e x)^{5/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{5/2}}-\frac {7 e (4 b B d-9 A b e+5 a B e)}{12 (b d-a e)^4 (d+e x)^{3/2}}-\frac {7 b e (4 b B d-9 A b e+5 a B e)}{4 (b d-a e)^5 \sqrt {d+e x}}-\frac {\left (7 b^2 (4 b B d-9 A b e+5 a B e)\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 (b d-a e)^5}\\ &=-\frac {7 e (4 b B d-9 A b e+5 a B e)}{20 b (b d-a e)^3 (d+e x)^{5/2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac {4 b B d-9 A b e+5 a B e}{4 b (b d-a e)^2 (a+b x) (d+e x)^{5/2}}-\frac {7 e (4 b B d-9 A b e+5 a B e)}{12 (b d-a e)^4 (d+e x)^{3/2}}-\frac {7 b e (4 b B d-9 A b e+5 a B e)}{4 (b d-a e)^5 \sqrt {d+e x}}+\frac {7 b^{3/2} e (4 b B d-9 A b e+5 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 97, normalized size = 0.35 \[ \frac {\frac {e (-5 a B e+9 A b e-4 b B d) \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+\frac {5 a B-5 A b}{(a+b x)^2}}{10 b (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)^3*(d + e*x)^(7/2)),x]

[Out]

((-5*A*b + 5*a*B)/(a + b*x)^2 + (e*(-4*b*B*d + 9*A*b*e - 5*a*B*e)*Hypergeometric2F1[-5/2, 2, -3/2, (b*(d + e*x
))/(b*d - a*e)])/(b*d - a*e)^2)/(10*b*(b*d - a*e)*(d + e*x)^(5/2))

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fricas [B]  time = 0.97, size = 2675, normalized size = 9.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

[1/120*(105*(4*B*a^2*b^2*d^4*e + (5*B*a^3*b - 9*A*a^2*b^2)*d^3*e^2 + (4*B*b^4*d*e^4 + (5*B*a*b^3 - 9*A*b^4)*e^
5)*x^5 + (12*B*b^4*d^2*e^3 + (23*B*a*b^3 - 27*A*b^4)*d*e^4 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*e^5)*x^4 + (12*B*b^4*
d^3*e^2 + 3*(13*B*a*b^3 - 9*A*b^4)*d^2*e^3 + 2*(17*B*a^2*b^2 - 27*A*a*b^3)*d*e^4 + (5*B*a^3*b - 9*A*a^2*b^2)*e
^5)*x^3 + (4*B*b^4*d^4*e + (29*B*a*b^3 - 9*A*b^4)*d^3*e^2 + 6*(7*B*a^2*b^2 - 9*A*a*b^3)*d^2*e^3 + 3*(5*B*a^3*b
- 9*A*a^2*b^2)*d*e^4)*x^2 + (8*B*a*b^3*d^4*e + 2*(11*B*a^2*b^2 - 9*A*a*b^3)*d^3*e^2 + 3*(5*B*a^3*b - 9*A*a^2*
b^2)*d^2*e^3)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)
))/(b*x + a)) + 2*(24*A*a^4*e^4 - 30*(B*a*b^3 + A*b^4)*d^4 - (659*B*a^2*b^2 - 255*A*a*b^3)*d^3*e - 16*(17*B*a^
3*b - 54*A*a^2*b^2)*d^2*e^2 + 8*(2*B*a^4 - 21*A*a^3*b)*d*e^3 - 105*(4*B*b^4*d*e^3 + (5*B*a*b^3 - 9*A*b^4)*e^4)
*x^4 - 35*(28*B*b^4*d^2*e^2 + (55*B*a*b^3 - 63*A*b^4)*d*e^3 + 5*(5*B*a^2*b^2 - 9*A*a*b^3)*e^4)*x^3 - 7*(92*B*b
^4*d^3*e + 9*(39*B*a*b^3 - 23*A*b^4)*d^2*e^2 + 3*(109*B*a^2*b^2 - 177*A*a*b^3)*d*e^3 + 8*(5*B*a^3*b - 9*A*a^2*
b^2)*e^4)*x^2 - (60*B*b^4*d^4 + (1183*B*a*b^3 - 135*A*b^4)*d^3*e + 3*(643*B*a^2*b^2 - 831*A*a*b^3)*d^2*e^2 + 7
2*(9*B*a^3*b - 17*A*a^2*b^2)*d*e^3 - 8*(5*B*a^4 - 9*A*a^3*b)*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d
^7*e + 10*a^4*b^3*d^6*e^2 - 10*a^5*b^2*d^5*e^3 + 5*a^6*b*d^4*e^4 - a^7*d^3*e^5 + (b^7*d^5*e^3 - 5*a*b^6*d^4*e^
4 + 10*a^2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6 + 5*a^4*b^3*d*e^7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d
^5*e^3 + 20*a^2*b^5*d^4*e^4 - 10*a^3*b^4*d^3*e^5 - 5*a^4*b^3*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2*a^6*b*e^8)*x^4 + (3
*b^7*d^7*e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*e^3 + 25*a^3*b^4*d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6
- a^6*b*d*e^7 - a^7*e^8)*x^3 + (b^7*d^8 + a*b^6*d^7*e - 17*a^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e^3 - 25*a^4*b^3*d
^4*e^4 - a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6 - 3*a^7*d*e^7)*x^2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6
*e^2 + 10*a^4*b^3*d^5*e^3 - 20*a^5*b^2*d^4*e^4 + 13*a^6*b*d^3*e^5 - 3*a^7*d^2*e^6)*x), 1/60*(105*(4*B*a^2*b^2*
d^4*e + (5*B*a^3*b - 9*A*a^2*b^2)*d^3*e^2 + (4*B*b^4*d*e^4 + (5*B*a*b^3 - 9*A*b^4)*e^5)*x^5 + (12*B*b^4*d^2*e^
3 + (23*B*a*b^3 - 27*A*b^4)*d*e^4 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*e^5)*x^4 + (12*B*b^4*d^3*e^2 + 3*(13*B*a*b^3 -
9*A*b^4)*d^2*e^3 + 2*(17*B*a^2*b^2 - 27*A*a*b^3)*d*e^4 + (5*B*a^3*b - 9*A*a^2*b^2)*e^5)*x^3 + (4*B*b^4*d^4*e
+ (29*B*a*b^3 - 9*A*b^4)*d^3*e^2 + 6*(7*B*a^2*b^2 - 9*A*a*b^3)*d^2*e^3 + 3*(5*B*a^3*b - 9*A*a^2*b^2)*d*e^4)*x^
2 + (8*B*a*b^3*d^4*e + 2*(11*B*a^2*b^2 - 9*A*a*b^3)*d^3*e^2 + 3*(5*B*a^3*b - 9*A*a^2*b^2)*d^2*e^3)*x)*sqrt(-b/
(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) + (24*A*a^4*e^4 - 30*(B*a*b
^3 + A*b^4)*d^4 - (659*B*a^2*b^2 - 255*A*a*b^3)*d^3*e - 16*(17*B*a^3*b - 54*A*a^2*b^2)*d^2*e^2 + 8*(2*B*a^4 -
21*A*a^3*b)*d*e^3 - 105*(4*B*b^4*d*e^3 + (5*B*a*b^3 - 9*A*b^4)*e^4)*x^4 - 35*(28*B*b^4*d^2*e^2 + (55*B*a*b^3 -
63*A*b^4)*d*e^3 + 5*(5*B*a^2*b^2 - 9*A*a*b^3)*e^4)*x^3 - 7*(92*B*b^4*d^3*e + 9*(39*B*a*b^3 - 23*A*b^4)*d^2*e^
2 + 3*(109*B*a^2*b^2 - 177*A*a*b^3)*d*e^3 + 8*(5*B*a^3*b - 9*A*a^2*b^2)*e^4)*x^2 - (60*B*b^4*d^4 + (1183*B*a*b
^3 - 135*A*b^4)*d^3*e + 3*(643*B*a^2*b^2 - 831*A*a*b^3)*d^2*e^2 + 72*(9*B*a^3*b - 17*A*a^2*b^2)*d*e^3 - 8*(5*B
*a^4 - 9*A*a^3*b)*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d^7*e + 10*a^4*b^3*d^6*e^2 - 10*a^5*b^2*d^5*
e^3 + 5*a^6*b*d^4*e^4 - a^7*d^3*e^5 + (b^7*d^5*e^3 - 5*a*b^6*d^4*e^4 + 10*a^2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6
+ 5*a^4*b^3*d*e^7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d^5*e^3 + 20*a^2*b^5*d^4*e^4 - 10*a^3*b^4*d^
3*e^5 - 5*a^4*b^3*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2*a^6*b*e^8)*x^4 + (3*b^7*d^7*e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*
e^3 + 25*a^3*b^4*d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6 - a^6*b*d*e^7 - a^7*e^8)*x^3 + (b^7*d^8 + a
*b^6*d^7*e - 17*a^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e^3 - 25*a^4*b^3*d^4*e^4 - a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6
- 3*a^7*d*e^7)*x^2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6*e^2 + 10*a^4*b^3*d^5*e^3 - 20*a^5*b^2*d^4*
e^4 + 13*a^6*b*d^3*e^5 - 3*a^7*d^2*e^6)*x)]

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giac [B]  time = 1.42, size = 609, normalized size = 2.17 \[ -\frac {7 \, {\left (4 \, B b^{3} d e + 5 \, B a b^{2} e^{2} - 9 \, A b^{3} e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt {-b^{2} d + a b e}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{4} d e - 4 \, \sqrt {x e + d} B b^{4} d^{2} e + 11 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{3} e^{2} - 15 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{4} e^{2} - 9 \, \sqrt {x e + d} B a b^{3} d e^{2} + 17 \, \sqrt {x e + d} A b^{4} d e^{2} + 13 \, \sqrt {x e + d} B a^{2} b^{2} e^{3} - 17 \, \sqrt {x e + d} A a b^{3} e^{3}}{4 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} - \frac {2 \, {\left (45 \, {\left (x e + d\right )}^{2} B b^{2} d e + 10 \, {\left (x e + d\right )} B b^{2} d^{2} e + 3 \, B b^{2} d^{3} e + 45 \, {\left (x e + d\right )}^{2} B a b e^{2} - 90 \, {\left (x e + d\right )}^{2} A b^{2} e^{2} - 5 \, {\left (x e + d\right )} B a b d e^{2} - 15 \, {\left (x e + d\right )} A b^{2} d e^{2} - 6 \, B a b d^{2} e^{2} - 3 \, A b^{2} d^{2} e^{2} - 5 \, {\left (x e + d\right )} B a^{2} e^{3} + 15 \, {\left (x e + d\right )} A a b e^{3} + 3 \, B a^{2} d e^{3} + 6 \, A a b d e^{3} - 3 \, A a^{2} e^{4}\right )}}{15 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-7/4*(4*B*b^3*d*e + 5*B*a*b^2*e^2 - 9*A*b^3*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^5 - 5*a*
b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*sqrt(-b^2*d + a*b*e)) - 1/4*(4*
(x*e + d)^(3/2)*B*b^4*d*e - 4*sqrt(x*e + d)*B*b^4*d^2*e + 11*(x*e + d)^(3/2)*B*a*b^3*e^2 - 15*(x*e + d)^(3/2)*
A*b^4*e^2 - 9*sqrt(x*e + d)*B*a*b^3*d*e^2 + 17*sqrt(x*e + d)*A*b^4*d*e^2 + 13*sqrt(x*e + d)*B*a^2*b^2*e^3 - 17
*sqrt(x*e + d)*A*a*b^3*e^3)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^
4 - a^5*e^5)*((x*e + d)*b - b*d + a*e)^2) - 2/15*(45*(x*e + d)^2*B*b^2*d*e + 10*(x*e + d)*B*b^2*d^2*e + 3*B*b^
2*d^3*e + 45*(x*e + d)^2*B*a*b*e^2 - 90*(x*e + d)^2*A*b^2*e^2 - 5*(x*e + d)*B*a*b*d*e^2 - 15*(x*e + d)*A*b^2*d
*e^2 - 6*B*a*b*d^2*e^2 - 3*A*b^2*d^2*e^2 - 5*(x*e + d)*B*a^2*e^3 + 15*(x*e + d)*A*a*b*e^3 + 3*B*a^2*d*e^3 + 6*
A*a*b*d*e^3 - 3*A*a^2*e^4)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4
- a^5*e^5)*(x*e + d)^(5/2))

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maple [B]  time = 0.03, size = 648, normalized size = 2.31 \[ -\frac {17 \sqrt {e x +d}\, A a \,b^{3} e^{3}}{4 \left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}+\frac {17 \sqrt {e x +d}\, A \,b^{4} d \,e^{2}}{4 \left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}+\frac {13 \sqrt {e x +d}\, B \,a^{2} b^{2} e^{3}}{4 \left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}-\frac {9 \sqrt {e x +d}\, B a \,b^{3} d \,e^{2}}{4 \left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}-\frac {\sqrt {e x +d}\, B \,b^{4} d^{2} e}{\left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}-\frac {15 \left (e x +d \right )^{\frac {3}{2}} A \,b^{4} e^{2}}{4 \left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}-\frac {63 A \,b^{3} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right )^{5} \sqrt {\left (a e -b d \right ) b}}+\frac {11 \left (e x +d \right )^{\frac {3}{2}} B a \,b^{3} e^{2}}{4 \left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}+\frac {35 B a \,b^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right )^{5} \sqrt {\left (a e -b d \right ) b}}+\frac {\left (e x +d \right )^{\frac {3}{2}} B \,b^{4} d e}{\left (a e -b d \right )^{5} \left (b x e +a e \right )^{2}}+\frac {7 B \,b^{3} d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5} \sqrt {\left (a e -b d \right ) b}}-\frac {12 A \,b^{2} e^{2}}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {6 B a b \,e^{2}}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {6 B \,b^{2} d e}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {2 A b \,e^{2}}{\left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 B a \,e^{2}}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {4 B b d e}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 A \,e^{2}}{5 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}}}+\frac {2 B d e}{5 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^3/(e*x+d)^(7/2),x)

[Out]

-15/4/(a*e-b*d)^5*b^4/(b*e*x+a*e)^2*(e*x+d)^(3/2)*A*e^2+11/4/(a*e-b*d)^5*b^3/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*a*e
^2+e/(a*e-b*d)^5*b^4/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*d-17/4/(a*e-b*d)^5*b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*a*e^3+
17/4/(a*e-b*d)^5*b^4/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*d*e^2+13/4/(a*e-b*d)^5*b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a^
2*e^3-9/4/(a*e-b*d)^5*b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a*d*e^2-e/(a*e-b*d)^5*b^4/(b*e*x+a*e)^2*(e*x+d)^(1/2)*
B*d^2-63/4/(a*e-b*d)^5*b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*e^2+35/4/(a*e-b*d
)^5*b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a*e^2+7*e/(a*e-b*d)^5*b^3/((a*e-b*d)
*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d-2/5/(a*e-b*d)^3/(e*x+d)^(5/2)*A*e^2+2/5*e/(a*e-b*d)^
3/(e*x+d)^(5/2)*B*d+2/(a*e-b*d)^4/(e*x+d)^(3/2)*A*b*e^2-2/3/(a*e-b*d)^4/(e*x+d)^(3/2)*B*a*e^2-4/3*e/(a*e-b*d)^
4/(e*x+d)^(3/2)*B*b*d-12*b^2/(a*e-b*d)^5/(e*x+d)^(1/2)*A*e^2+6*b/(a*e-b*d)^5/(e*x+d)^(1/2)*B*a*e^2+6*e*b^2/(a*
e-b*d)^5/(e*x+d)^(1/2)*B*d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(7/2),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 1.65, size = 418, normalized size = 1.49 \[ \frac {\frac {35\,{\left (d+e\,x\right )}^3\,\left (-9\,A\,b^3\,e^2+4\,B\,d\,b^3\,e+5\,B\,a\,b^2\,e^2\right )}{12\,{\left (a\,e-b\,d\right )}^4}-\frac {2\,\left (A\,e^2-B\,d\,e\right )}{5\,\left (a\,e-b\,d\right )}-\frac {2\,\left (d+e\,x\right )\,\left (5\,B\,a\,e^2-9\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{15\,{\left (a\,e-b\,d\right )}^2}+\frac {7\,b^3\,{\left (d+e\,x\right )}^4\,\left (5\,B\,a\,e^2-9\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^5}+\frac {14\,b\,{\left (d+e\,x\right )}^2\,\left (5\,B\,a\,e^2-9\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{15\,{\left (a\,e-b\,d\right )}^3}}{b^2\,{\left (d+e\,x\right )}^{9/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{7/2}+{\left (d+e\,x\right )}^{5/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}+\frac {7\,b^{3/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (5\,B\,a\,e-9\,A\,b\,e+4\,B\,b\,d\right )\,\left (a^5\,e^5-5\,a^4\,b\,d\,e^4+10\,a^3\,b^2\,d^2\,e^3-10\,a^2\,b^3\,d^3\,e^2+5\,a\,b^4\,d^4\,e-b^5\,d^5\right )}{{\left (a\,e-b\,d\right )}^{11/2}\,\left (5\,B\,a\,e^2-9\,A\,b\,e^2+4\,B\,b\,d\,e\right )}\right )\,\left (5\,B\,a\,e-9\,A\,b\,e+4\,B\,b\,d\right )}{4\,{\left (a\,e-b\,d\right )}^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)^3*(d + e*x)^(7/2)),x)

[Out]

((35*(d + e*x)^3*(4*B*b^3*d*e - 9*A*b^3*e^2 + 5*B*a*b^2*e^2))/(12*(a*e - b*d)^4) - (2*(A*e^2 - B*d*e))/(5*(a*e
- b*d)) - (2*(d + e*x)*(5*B*a*e^2 - 9*A*b*e^2 + 4*B*b*d*e))/(15*(a*e - b*d)^2) + (7*b^3*(d + e*x)^4*(5*B*a*e^
2 - 9*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^5) + (14*b*(d + e*x)^2*(5*B*a*e^2 - 9*A*b*e^2 + 4*B*b*d*e))/(15*(a*
e - b*d)^3))/(b^2*(d + e*x)^(9/2) - (2*b^2*d - 2*a*b*e)*(d + e*x)^(7/2) + (d + e*x)^(5/2)*(a^2*e^2 + b^2*d^2 -
2*a*b*d*e)) + (7*b^(3/2)*e*atan((b^(1/2)*e*(d + e*x)^(1/2)*(5*B*a*e - 9*A*b*e + 4*B*b*d)*(a^5*e^5 - b^5*d^5 -
10*a^2*b^3*d^3*e^2 + 10*a^3*b^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4))/((a*e - b*d)^(11/2)*(5*B*a*e^2 - 9*
A*b*e^2 + 4*B*b*d*e)))*(5*B*a*e - 9*A*b*e + 4*B*b*d))/(4*(a*e - b*d)^(11/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)**3/(e*x+d)**(7/2),x)

[Out]

Timed out

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