3.1815 \(\int \frac {A+B x}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=221 \[ -\frac {b^{3/2} (5 a B e-7 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{9/2}}+\frac {b (5 a B e-7 A b e+2 b B d)}{\sqrt {d+e x} (b d-a e)^4}+\frac {5 a B e-7 A b e+2 b B d}{3 (d+e x)^{3/2} (b d-a e)^3}+\frac {5 a B e-7 A b e+2 b B d}{5 b (d+e x)^{5/2} (b d-a e)^2}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 78, 51, 63, 208} \[ -\frac {b^{3/2} (5 a B e-7 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{9/2}}+\frac {b (5 a B e-7 A b e+2 b B d)}{\sqrt {d+e x} (b d-a e)^4}+\frac {5 a B e-7 A b e+2 b B d}{3 (d+e x)^{3/2} (b d-a e)^3}+\frac {5 a B e-7 A b e+2 b B d}{5 b (d+e x)^{5/2} (b d-a e)^2}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

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Mathematica [C]  time = 0.04, size = 94, normalized size = 0.43 \[ \frac {(5 a B e-7 A b e+2 b B d) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )-\frac {5 (A b-a B) (b d-a e)}{a+b x}}{5 b (d+e x)^{5/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.85, size = 1749, normalized size = 7.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30*(15*(2*B*a*b^2*d^4 + (5*B*a^2*b - 7*A*a*b^2)*d^3*e + (2*B*b^3*d*e^3 + (5*B*a*b^2 - 7*A*b^3)*e^4)*x^4 +
(6*B*b^3*d^2*e^2 + (17*B*a*b^2 - 21*A*b^3)*d*e^3 + (5*B*a^2*b - 7*A*a*b^2)*e^4)*x^3 + 3*(2*B*b^3*d^3*e + 7*(B*
a*b^2 - A*b^3)*d^2*e^2 + (5*B*a^2*b - 7*A*a*b^2)*d*e^3)*x^2 + (2*B*b^3*d^4 + (11*B*a*b^2 - 7*A*b^3)*d^3*e + 3*
(5*B*a^2*b - 7*A*a*b^2)*d^2*e^2)*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*(6*A*a^3*e^3 - (61*B*a*b^2 - 15*A*b^3)*d^3 - 4*(12*B*a^2*b - 29*A*a*b^2)*
d^2*e + 4*(B*a^3 - 8*A*a^2*b)*d*e^2 - 15*(2*B*b^3*d*e^2 + (5*B*a*b^2 - 7*A*b^3)*e^3)*x^3 - 5*(14*B*b^3*d^2*e +
(39*B*a*b^2 - 49*A*b^3)*d*e^2 + 2*(5*B*a^2*b - 7*A*a*b^2)*e^3)*x^2 - (46*B*b^3*d^3 + (163*B*a*b^2 - 161*A*b^3
)*d^2*e + 4*(29*B*a^2*b - 42*A*a*b^2)*d*e^2 - 2*(5*B*a^3 - 7*A*a^2*b)*e^3)*x)*sqrt(e*x + d))/(a*b^4*d^7 - 4*a^
2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4*a*b^4*d^3*e^4 + 6*a^2*b^3*d
^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e^3 + 14*a^2*b^3*d^3*e^4 - 6*a^3*b^2
*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*a^2*b^3*d^4*e^3 + 2*a^3*b^2*d^3*e^4
- 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^5*e^2 + 14*a^3*b^2*d^4*e^3 - 11*a^4
*b*d^3*e^4 + 3*a^5*d^2*e^5)*x), -1/15*(15*(2*B*a*b^2*d^4 + (5*B*a^2*b - 7*A*a*b^2)*d^3*e + (2*B*b^3*d*e^3 + (5
*B*a*b^2 - 7*A*b^3)*e^4)*x^4 + (6*B*b^3*d^2*e^2 + (17*B*a*b^2 - 21*A*b^3)*d*e^3 + (5*B*a^2*b - 7*A*a*b^2)*e^4)
*x^3 + 3*(2*B*b^3*d^3*e + 7*(B*a*b^2 - A*b^3)*d^2*e^2 + (5*B*a^2*b - 7*A*a*b^2)*d*e^3)*x^2 + (2*B*b^3*d^4 + (1
1*B*a*b^2 - 7*A*b^3)*d^3*e + 3*(5*B*a^2*b - 7*A*a*b^2)*d^2*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sq
rt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) + (6*A*a^3*e^3 - (61*B*a*b^2 - 15*A*b^3)*d^3 - 4*(12*B*a^2*b -
29*A*a*b^2)*d^2*e + 4*(B*a^3 - 8*A*a^2*b)*d*e^2 - 15*(2*B*b^3*d*e^2 + (5*B*a*b^2 - 7*A*b^3)*e^3)*x^3 - 5*(14*
B*b^3*d^2*e + (39*B*a*b^2 - 49*A*b^3)*d*e^2 + 2*(5*B*a^2*b - 7*A*a*b^2)*e^3)*x^2 - (46*B*b^3*d^3 + (163*B*a*b^
2 - 161*A*b^3)*d^2*e + 4*(29*B*a^2*b - 42*A*a*b^2)*d*e^2 - 2*(5*B*a^3 - 7*A*a^2*b)*e^3)*x)*sqrt(e*x + d))/(a*b
^4*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4*a*b^4*d^3*e^4
+ 6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e^3 + 14*a^2*b^3*d^3*e^
4 - 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*a^2*b^3*d^4*e^3 + 2*a^
3*b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^5*e^2 + 14*a^3*b^2*d^4
*e^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x)]

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giac [B]  time = 0.23, size = 435, normalized size = 1.97 \[ \frac {{\left (2 \, B b^{3} d + 5 \, B a b^{2} e - 7 \, A b^{3} e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e}} + \frac {\sqrt {x e + d} B a b^{2} e - \sqrt {x e + d} A b^{3} e}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}} + \frac {2 \, {\left (15 \, {\left (x e + d\right )}^{2} B b^{2} d + 5 \, {\left (x e + d\right )} B b^{2} d^{2} + 3 \, B b^{2} d^{3} + 30 \, {\left (x e + d\right )}^{2} B a b e - 45 \, {\left (x e + d\right )}^{2} A b^{2} e - 10 \, {\left (x e + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (x e + d\right )} B a^{2} e^{2} + 10 \, {\left (x e + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )}}{15 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*b^3*d + 5*B*a*b^2*e - 7*A*b^3*e)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^4 - 4*a*b^3*d^3*e +
6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*sqrt(-b^2*d + a*b*e)) + (sqrt(x*e + d)*B*a*b^2*e - sqrt(x*e + d)
*A*b^3*e)/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*((x*e + d)*b - b*d + a*e))
+ 2/15*(15*(x*e + d)^2*B*b^2*d + 5*(x*e + d)*B*b^2*d^2 + 3*B*b^2*d^3 + 30*(x*e + d)^2*B*a*b*e - 45*(x*e + d)^2
*A*b^2*e - 10*(x*e + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A*b^2*d^2*e - 5*(x*e + d)*B*a^2*e^2 + 10*(x*e + d)*A*a*b
*e^2 + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 - 3*A*a^2*e^3)/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*
e^3 + a^4*e^4)*(x*e + d)^(5/2))

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maple [B]  time = 0.07, size = 403, normalized size = 1.82 \[ -\frac {7 A \,b^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{4} \sqrt {\left (a e -b d \right ) b}}+\frac {5 B a \,b^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{4} \sqrt {\left (a e -b d \right ) b}}+\frac {2 B \,b^{3} d \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{4} \sqrt {\left (a e -b d \right ) b}}-\frac {\sqrt {e x +d}\, A \,b^{3} e}{\left (a e -b d \right )^{4} \left (b e x +a e \right )}+\frac {\sqrt {e x +d}\, B a \,b^{2} e}{\left (a e -b d \right )^{4} \left (b e x +a e \right )}-\frac {6 A \,b^{2} e}{\left (a e -b d \right )^{4} \sqrt {e x +d}}+\frac {4 B a b e}{\left (a e -b d \right )^{4} \sqrt {e x +d}}+\frac {2 B \,b^{2} d}{\left (a e -b d \right )^{4} \sqrt {e x +d}}+\frac {4 A b e}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 B a e}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 B b d}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 A e}{5 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {5}{2}}}+\frac {2 B d}{5 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*e-b*d)^4*b^3*(e*x+d)^(1/2)/(b*e*x+a*e)*A*e+1/(a*e-b*d)^4*b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*a*B*e-7/(a*e-b*d)
^4*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+5/(a*e-b*d)^4*b^2/((a*e-b*d)*b)^(1/
2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*B*e+2/(a*e-b*d)^4*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)^2/(e*x+d)^(5/2)*A*e+2/5/(a*e-b*d)^2/(e*x+d)^(5/2)*B*d+4/3/(a*e-b*d)
^3/(e*x+d)^(3/2)*A*b*e-2/3/(a*e-b*d)^3/(e*x+d)^(3/2)*a*B*e-2/3/(a*e-b*d)^3/(e*x+d)^(3/2)*B*b*d-6*b^2/(a*e-b*d)
^4/(e*x+d)^(1/2)*A*e+4*b/(a*e-b*d)^4/(e*x+d)^(1/2)*a*B*e+2*b^2/(a*e-b*d)^4/(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)/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^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 = 2.16, size = 261, normalized size = 1.18 \[ \frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^{9/2}}\right )\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^{9/2}}-\frac {\frac {2\,\left (A\,e-B\,d\right )}{5\,\left (a\,e-b\,d\right )}+\frac {2\,\left (d+e\,x\right )\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{15\,{\left (a\,e-b\,d\right )}^2}-\frac {b^2\,{\left (d+e\,x\right )}^3\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^4}-\frac {2\,b\,{\left (d+e\,x\right )}^2\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^3}}{b\,{\left (d+e\,x\right )}^{7/2}+\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^(3/2)*atan((b^(1/2)*(d + e*x)^(1/2)*(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3)
)/(a*e - b*d)^(9/2))*(5*B*a*e - 7*A*b*e + 2*B*b*d))/(a*e - b*d)^(9/2) - ((2*(A*e - B*d))/(5*(a*e - b*d)) + (2*
(d + e*x)*(5*B*a*e - 7*A*b*e + 2*B*b*d))/(15*(a*e - b*d)^2) - (b^2*(d + e*x)^3*(5*B*a*e - 7*A*b*e + 2*B*b*d))/
(a*e - b*d)^4 - (2*b*(d + e*x)^2*(5*B*a*e - 7*A*b*e + 2*B*b*d))/(3*(a*e - b*d)^3))/(b*(d + e*x)^(7/2) + (a*e -
b*d)*(d + e*x)^(5/2))

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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)/(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Timed out

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