3.1761 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^3} \, dx\)

Optimal. Leaf size=157 \[ -\frac {e (-3 a B e-A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x} (-3 a B e-A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}-\frac {(d+e x)^{3/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 47, 63, 208} \[ -\frac {\sqrt {d+e x} (-3 a B e-A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}-\frac {e (-3 a B e-A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} (b d-a e)^{3/2}}-\frac {(d+e x)^{3/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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) \sqrt {d+e x}}{(a+b x)^3} \, dx &=-\frac {(A b-a B) (d+e x)^{3/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(4 b B d-A b e-3 a B e) \int \frac {\sqrt {d+e x}}{(a+b x)^2} \, dx}{4 b (b d-a e)}\\ &=-\frac {(4 b B d-A b e-3 a B e) \sqrt {d+e x}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{3/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(e (4 b B d-A b e-3 a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^2 (b d-a e)}\\ &=-\frac {(4 b B d-A b e-3 a B e) \sqrt {d+e x}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{3/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(4 b B d-A b e-3 a B e) \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^2 (b d-a e)}\\ &=-\frac {(4 b B d-A b e-3 a B e) \sqrt {d+e x}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{3/2}}{2 b (b d-a e) (a+b x)^2}-\frac {e (4 b B d-A b e-3 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} (b d-a e)^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.37, size = 158, normalized size = 1.01 \[ \frac {\frac {(a+b x) (-3 a B e-A b e+4 b B d) \left (\sqrt {b} e (a+b x) \sqrt {d+e x} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )-b (d+e x) \sqrt {a e-b d}\right )}{\sqrt {a e-b d}}-2 b^2 (d+e x)^2 (A b-a B)}{4 b^3 (a+b x)^2 \sqrt {d+e x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.89, size = 721, normalized size = 4.59 \[ \left [\frac {{\left (4 \, B a^{2} b d e - {\left (3 \, B a^{3} + A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (3 \, B a^{2} b + A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (5 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} d e + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (9 \, B a b^{3} - A b^{4}\right )} d e + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{5} d^{2} - 2 \, a^{3} b^{4} d e + a^{4} b^{3} e^{2} + {\left (b^{7} d^{2} - 2 \, a b^{6} d e + a^{2} b^{5} e^{2}\right )} x^{2} + 2 \, {\left (a b^{6} d^{2} - 2 \, a^{2} b^{5} d e + a^{3} b^{4} e^{2}\right )} x\right )}}, \frac {{\left (4 \, B a^{2} b d e - {\left (3 \, B a^{3} + A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (3 \, B a^{2} b + A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (5 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} d e + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (9 \, B a b^{3} - A b^{4}\right )} d e + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{5} d^{2} - 2 \, a^{3} b^{4} d e + a^{4} b^{3} e^{2} + {\left (b^{7} d^{2} - 2 \, a b^{6} d e + a^{2} b^{5} e^{2}\right )} x^{2} + 2 \, {\left (a b^{6} d^{2} - 2 \, a^{2} b^{5} d e + a^{3} b^{4} e^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[1/8*((4*B*a^2*b*d*e - (3*B*a^3 + A*a^2*b)*e^2 + (4*B*b^3*d*e - (3*B*a*b^2 + A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*
e - (3*B*a^2*b + A*a*b^2)*e^2)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*
x + d))/(b*x + a)) - 2*(2*(B*a*b^3 + A*b^4)*d^2 - (5*B*a^2*b^2 + 3*A*a*b^3)*d*e + (3*B*a^3*b + A*a^2*b^2)*e^2
+ (4*B*b^4*d^2 - (9*B*a*b^3 - A*b^4)*d*e + (5*B*a^2*b^2 - A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^2 - 2*a^3
*b^4*d*e + a^4*b^3*e^2 + (b^7*d^2 - 2*a*b^6*d*e + a^2*b^5*e^2)*x^2 + 2*(a*b^6*d^2 - 2*a^2*b^5*d*e + a^3*b^4*e^
2)*x), 1/4*((4*B*a^2*b*d*e - (3*B*a^3 + A*a^2*b)*e^2 + (4*B*b^3*d*e - (3*B*a*b^2 + A*b^3)*e^2)*x^2 + 2*(4*B*a*
b^2*d*e - (3*B*a^2*b + A*a*b^2)*e^2)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x
+ b*d)) - (2*(B*a*b^3 + A*b^4)*d^2 - (5*B*a^2*b^2 + 3*A*a*b^3)*d*e + (3*B*a^3*b + A*a^2*b^2)*e^2 + (4*B*b^4*d^
2 - (9*B*a*b^3 - A*b^4)*d*e + (5*B*a^2*b^2 - A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^2 - 2*a^3*b^4*d*e + a^
4*b^3*e^2 + (b^7*d^2 - 2*a*b^6*d*e + a^2*b^5*e^2)*x^2 + 2*(a*b^6*d^2 - 2*a^2*b^5*d*e + a^3*b^4*e^2)*x)]

________________________________________________________________________________________

giac [A]  time = 1.24, size = 245, normalized size = 1.56 \[ \frac {{\left (4 \, B b d e - 3 \, B a e^{2} - A b e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{3} d - a b^{2} e\right )} \sqrt {-b^{2} d + a b e}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e - 4 \, \sqrt {x e + d} B b^{2} d^{2} e - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{2} + {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{2} + 7 \, \sqrt {x e + d} B a b d e^{2} + \sqrt {x e + d} A b^{2} d e^{2} - 3 \, \sqrt {x e + d} B a^{2} e^{3} - \sqrt {x e + d} A a b e^{3}}{4 \, {\left (b^{3} d - a b^{2} e\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 339, normalized size = 2.16 \[ \frac {A \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} A \,e^{2}}{4 \left (b x e +a e \right )^{2} \left (a e -b d \right )}-\frac {5 \left (e x +d \right )^{\frac {3}{2}} B a \,e^{2}}{4 \left (b x e +a e \right )^{2} \left (a e -b d \right ) b}+\frac {3 B a \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {B d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} B d e}{\left (b x e +a e \right )^{2} \left (a e -b d \right )}-\frac {\sqrt {e x +d}\, A \,e^{2}}{4 \left (b x e +a e \right )^{2} b}-\frac {3 \sqrt {e x +d}\, B a \,e^{2}}{4 \left (b x e +a e \right )^{2} b^{2}}+\frac {\sqrt {e x +d}\, B d e}{\left (b x e +a e \right )^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^3,x)

[Out]

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

________________________________________________________________________________________

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)^(1/2)/(b*x+a)^3,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?

________________________________________________________________________________________

mupad [B]  time = 0.22, size = 222, normalized size = 1.41 \[ \frac {e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (A\,b\,e+3\,B\,a\,e-4\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^2+3\,B\,a\,e^2-4\,B\,b\,d\,e\right )}\right )\,\left (A\,b\,e+3\,B\,a\,e-4\,B\,b\,d\right )}{4\,b^{5/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {\sqrt {d+e\,x}\,\left (A\,b\,e^2+3\,B\,a\,e^2-4\,B\,b\,d\,e\right )}{4\,b^2}-\frac {{\left (d+e\,x\right )}^{3/2}\,\left (A\,b\,e^2-5\,B\,a\,e^2+4\,B\,b\,d\,e\right )}{4\,b\,\left (a\,e-b\,d\right )}}{b^2\,{\left (d+e\,x\right )}^2-\left (2\,b^2\,d-2\,a\,b\,e\right )\,\left (d+e\,x\right )+a^2\,e^2+b^2\,d^2-2\,a\,b\,d\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(1/2))/(a + b*x)^3,x)

[Out]

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

________________________________________________________________________________________

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)**(1/2)/(b*x+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________