3.1751 \(\int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=214 \[ -\frac {(b d-a e)^{3/2} (-7 a B e+5 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {\sqrt {d+e x} (b d-a e) (-7 a B e+5 A b e+2 b B d)}{b^4}+\frac {(d+e x)^{3/2} (-7 a B e+5 A b e+2 b B d)}{3 b^3}+\frac {(d+e x)^{5/2} (-7 a B e+5 A b e+2 b B d)}{5 b^2 (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 214, 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, 50, 63, 208} \[ \frac {(d+e x)^{5/2} (-7 a B e+5 A b e+2 b B d)}{5 b^2 (b d-a e)}+\frac {(d+e x)^{3/2} (-7 a B e+5 A b e+2 b B d)}{3 b^3}+\frac {\sqrt {d+e x} (b d-a e) (-7 a B e+5 A b e+2 b B d)}{b^4}-\frac {(b d-a e)^{3/2} (-7 a B e+5 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(5/2))/(a + b*x)^2,x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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)^{5/2}}{(a+b x)^2} \, dx &=-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {(2 b B d+5 A b e-7 a B e) \int \frac {(d+e x)^{5/2}}{a+b x} \, dx}{2 b (b d-a e)}\\ &=\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {(2 b B d+5 A b e-7 a B e) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {((b d-a e) (2 b B d+5 A b e-7 a B e)) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{2 b^3}\\ &=\frac {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {\left ((b d-a e)^2 (2 b B d+5 A b e-7 a B e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 b^4}\\ &=\frac {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}+\frac {\left ((b d-a e)^2 (2 b B d+5 A b e-7 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 )}{b^4 e}\\ &=\frac {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}-\frac {(b d-a e)^{3/2} (2 b B d+5 A b e-7 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 166, normalized size = 0.78 \[ \frac {\frac {2 \left (-\frac {7 a B e}{2}+\frac {5 A b e}{2}+b B d\right ) \left (5 (b d-a e) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )+3 b^{5/2} (d+e x)^{5/2}\right )}{15 b^{7/2}}+\frac {(d+e x)^{7/2} (a B-A b)}{a+b x}}{b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-(A*b) + a*B)*(d + e*x)^(7/2))/(a + b*x) + (2*(b*B*d + (5*A*b*e)/2 - (7*a*B*e)/2)*(3*b^(5/2)*(d + e*x)^(5/2
) + 5*(b*d - a*e)*(Sqrt[b]*Sqrt[d + e*x]*(4*b*d - 3*a*e + b*e*x) - 3*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])))/(15*b^(7/2)))/(b*(b*d - a*e))

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fricas [A]  time = 0.78, size = 666, normalized size = 3.11 \[ \left [-\frac {15 \, {\left (2 \, B a b^{2} d^{2} - {\left (9 \, B a^{2} b - 5 \, A a b^{2}\right )} d e + {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d^{2} - {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (6 \, B b^{3} e^{2} x^{3} + {\left (61 \, B a b^{2} - 15 \, A b^{3}\right )} d^{2} - 10 \, {\left (17 \, B a^{2} b - 10 \, A a b^{2}\right )} d e + 15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + 2 \, {\left (11 \, B b^{3} d e - {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{3} d^{2} - {\left (59 \, B a b^{2} - 35 \, A b^{3}\right )} d e + 5 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, B a b^{2} d^{2} - {\left (9 \, B a^{2} b - 5 \, A a b^{2}\right )} d e + {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d^{2} - {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (6 \, B b^{3} e^{2} x^{3} + {\left (61 \, B a b^{2} - 15 \, A b^{3}\right )} d^{2} - 10 \, {\left (17 \, B a^{2} b - 10 \, A a b^{2}\right )} d e + 15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + 2 \, {\left (11 \, B b^{3} d e - {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{3} d^{2} - {\left (59 \, B a b^{2} - 35 \, A b^{3}\right )} d e + 5 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30*(15*(2*B*a*b^2*d^2 - (9*B*a^2*b - 5*A*a*b^2)*d*e + (7*B*a^3 - 5*A*a^2*b)*e^2 + (2*B*b^3*d^2 - (9*B*a*b^
2 - 5*A*b^3)*d*e + (7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e + 2*sqrt(e*x +
d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(6*B*b^3*e^2*x^3 + (61*B*a*b^2 - 15*A*b^3)*d^2 - 10*(17*B*a^2*b - 10
*A*a*b^2)*d*e + 15*(7*B*a^3 - 5*A*a^2*b)*e^2 + 2*(11*B*b^3*d*e - (7*B*a*b^2 - 5*A*b^3)*e^2)*x^2 + 2*(23*B*b^3*
d^2 - (59*B*a*b^2 - 35*A*b^3)*d*e + 5*(7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^5*x + a*b^4), -1/15*(1
5*(2*B*a*b^2*d^2 - (9*B*a^2*b - 5*A*a*b^2)*d*e + (7*B*a^3 - 5*A*a^2*b)*e^2 + (2*B*b^3*d^2 - (9*B*a*b^2 - 5*A*b
^3)*d*e + (7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b
*d - a*e)) - (6*B*b^3*e^2*x^3 + (61*B*a*b^2 - 15*A*b^3)*d^2 - 10*(17*B*a^2*b - 10*A*a*b^2)*d*e + 15*(7*B*a^3 -
5*A*a^2*b)*e^2 + 2*(11*B*b^3*d*e - (7*B*a*b^2 - 5*A*b^3)*e^2)*x^2 + 2*(23*B*b^3*d^2 - (59*B*a*b^2 - 35*A*b^3)
*d*e + 5*(7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^5*x + a*b^4)]

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giac [B]  time = 1.29, size = 400, normalized size = 1.87 \[ \frac {{\left (2 \, B b^{3} d^{3} - 11 \, B a b^{2} d^{2} e + 5 \, A b^{3} d^{2} e + 16 \, B a^{2} b d e^{2} - 10 \, A a b^{2} d e^{2} - 7 \, B a^{3} e^{3} + 5 \, A a^{2} b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{4}} + \frac {\sqrt {x e + d} B a b^{2} d^{2} e - \sqrt {x e + d} A b^{3} d^{2} e - 2 \, \sqrt {x e + d} B a^{2} b d e^{2} + 2 \, \sqrt {x e + d} A a b^{2} d e^{2} + \sqrt {x e + d} B a^{3} e^{3} - \sqrt {x e + d} A a^{2} b e^{3}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{8} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{8} d + 15 \, \sqrt {x e + d} B b^{8} d^{2} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{7} e + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{8} e - 60 \, \sqrt {x e + d} B a b^{7} d e + 30 \, \sqrt {x e + d} A b^{8} d e + 45 \, \sqrt {x e + d} B a^{2} b^{6} e^{2} - 30 \, \sqrt {x e + d} A a b^{7} e^{2}\right )}}{15 \, b^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*b^3*d^3 - 11*B*a*b^2*d^2*e + 5*A*b^3*d^2*e + 16*B*a^2*b*d*e^2 - 10*A*a*b^2*d*e^2 - 7*B*a^3*e^3 + 5*A*a^2*
b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4) + (sqrt(x*e + d)*B*a*b^2*d^2*e
- sqrt(x*e + d)*A*b^3*d^2*e - 2*sqrt(x*e + d)*B*a^2*b*d*e^2 + 2*sqrt(x*e + d)*A*a*b^2*d*e^2 + sqrt(x*e + d)*B*
a^3*e^3 - sqrt(x*e + d)*A*a^2*b*e^3)/(((x*e + d)*b - b*d + a*e)*b^4) + 2/15*(3*(x*e + d)^(5/2)*B*b^8 + 5*(x*e
+ d)^(3/2)*B*b^8*d + 15*sqrt(x*e + d)*B*b^8*d^2 - 10*(x*e + d)^(3/2)*B*a*b^7*e + 5*(x*e + d)^(3/2)*A*b^8*e - 6
0*sqrt(x*e + d)*B*a*b^7*d*e + 30*sqrt(x*e + d)*A*b^8*d*e + 45*sqrt(x*e + d)*B*a^2*b^6*e^2 - 30*sqrt(x*e + d)*A
*a*b^7*e^2)/b^10

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maple [B]  time = 0.02, size = 626, normalized size = 2.93 \[ \frac {5 A \,a^{2} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {10 A a d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {5 A \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}-\frac {7 B \,a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{4}}+\frac {16 B \,a^{2} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {11 B a \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {2 B \,d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}-\frac {\sqrt {e x +d}\, A \,a^{2} e^{3}}{\left (b x e +a e \right ) b^{3}}+\frac {2 \sqrt {e x +d}\, A a d \,e^{2}}{\left (b x e +a e \right ) b^{2}}-\frac {\sqrt {e x +d}\, A \,d^{2} e}{\left (b x e +a e \right ) b}+\frac {\sqrt {e x +d}\, B \,a^{3} e^{3}}{\left (b x e +a e \right ) b^{4}}-\frac {2 \sqrt {e x +d}\, B \,a^{2} d \,e^{2}}{\left (b x e +a e \right ) b^{3}}+\frac {\sqrt {e x +d}\, B a \,d^{2} e}{\left (b x e +a e \right ) b^{2}}-\frac {4 \sqrt {e x +d}\, A a \,e^{2}}{b^{3}}+\frac {4 \sqrt {e x +d}\, A d e}{b^{2}}+\frac {6 \sqrt {e x +d}\, B \,a^{2} e^{2}}{b^{4}}-\frac {8 \sqrt {e x +d}\, B a d e}{b^{3}}+\frac {2 \sqrt {e x +d}\, B \,d^{2}}{b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} A e}{3 b^{2}}-\frac {4 \left (e x +d \right )^{\frac {3}{2}} B a e}{3 b^{3}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B d}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} B}{5 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(5/2)/(b*x+a)^2,x)

[Out]

2/5/b^2*B*(e*x+d)^(5/2)+2/3/b^2*A*(e*x+d)^(3/2)*e-4/3/b^3*B*(e*x+d)^(3/2)*a*e+2/3/b^2*B*(e*x+d)^(3/2)*d-4/b^3*
A*(e*x+d)^(1/2)*a*e^2+4/b^2*A*(e*x+d)^(1/2)*d*e+6/b^4*B*(e*x+d)^(1/2)*a^2*e^2-8/b^3*B*(e*x+d)^(1/2)*a*d*e+2/b^
2*B*(e*x+d)^(1/2)*d^2-1/b^3*(e*x+d)^(1/2)/(b*e*x+a*e)*A*a^2*e^3+2/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*A*a*d*e^2-1/b*
(e*x+d)^(1/2)/(b*e*x+a*e)*A*d^2*e+1/b^4*(e*x+d)^(1/2)/(b*e*x+a*e)*B*a^3*e^3-2/b^3*(e*x+d)^(1/2)/(b*e*x+a*e)*B*
a^2*d*e^2+1/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*B*a*d^2*e+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*a^2*e^3-10/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*a*d*e^2+5/b/((a
*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*d^2*e-7/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^
(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a^3*e^3+16/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*
B*a^2*d*e^2-11/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*d^2*e+2/b/((a*e-b*d)*b)
^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d^3

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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)^(5/2)/(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 = 0.13, size = 363, normalized size = 1.70 \[ \left (\frac {2\,A\,e-2\,B\,d}{3\,b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{3\,b^4}\right )\,{\left (d+e\,x\right )}^{3/2}+\left (\frac {\left (2\,b^2\,d-2\,a\,b\,e\right )\,\left (\frac {2\,A\,e-2\,B\,d}{b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{b^4}\right )}{b^2}-\frac {2\,B\,{\left (a\,e-b\,d\right )}^2}{b^4}\right )\,\sqrt {d+e\,x}+\frac {\sqrt {d+e\,x}\,\left (B\,a^3\,e^3-2\,B\,a^2\,b\,d\,e^2-A\,a^2\,b\,e^3+B\,a\,b^2\,d^2\,e+2\,A\,a\,b^2\,d\,e^2-A\,b^3\,d^2\,e\right )}{b^5\,\left (d+e\,x\right )-b^5\,d+a\,b^4\,e}+\frac {2\,B\,{\left (d+e\,x\right )}^{5/2}}{5\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}\,\left (5\,A\,b\,e-7\,B\,a\,e+2\,B\,b\,d\right )}{-7\,B\,a^3\,e^3+16\,B\,a^2\,b\,d\,e^2+5\,A\,a^2\,b\,e^3-11\,B\,a\,b^2\,d^2\,e-10\,A\,a\,b^2\,d\,e^2+2\,B\,b^3\,d^3+5\,A\,b^3\,d^2\,e}\right )\,{\left (a\,e-b\,d\right )}^{3/2}\,\left (5\,A\,b\,e-7\,B\,a\,e+2\,B\,b\,d\right )}{b^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(5/2))/(a + b*x)^2,x)

[Out]

((2*A*e - 2*B*d)/(3*b^2) + (2*B*(2*b^2*d - 2*a*b*e))/(3*b^4))*(d + e*x)^(3/2) + (((2*b^2*d - 2*a*b*e)*((2*A*e
- 2*B*d)/b^2 + (2*B*(2*b^2*d - 2*a*b*e))/b^4))/b^2 - (2*B*(a*e - b*d)^2)/b^4)*(d + e*x)^(1/2) + ((d + e*x)^(1/
2)*(B*a^3*e^3 - A*a^2*b*e^3 - A*b^3*d^2*e + 2*A*a*b^2*d*e^2 + B*a*b^2*d^2*e - 2*B*a^2*b*d*e^2))/(b^5*(d + e*x)
- b^5*d + a*b^4*e) + (2*B*(d + e*x)^(5/2))/(5*b^2) + (atan((b^(1/2)*(a*e - b*d)^(3/2)*(d + e*x)^(1/2)*(5*A*b*
e - 7*B*a*e + 2*B*b*d))/(2*B*b^3*d^3 - 7*B*a^3*e^3 + 5*A*a^2*b*e^3 + 5*A*b^3*d^2*e - 10*A*a*b^2*d*e^2 - 11*B*a
*b^2*d^2*e + 16*B*a^2*b*d*e^2))*(a*e - b*d)^(3/2)*(5*A*b*e - 7*B*a*e + 2*B*b*d))/b^(9/2)

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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)**(5/2)/(b*x+a)**2,x)

[Out]

Timed out

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