3.18.61 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^3} \, dx\) [1761]

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

[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
)

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Rubi [A]
time = 0.08, 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 = {79, 43, 65, 214} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((4*b*B*d - A*b*e - 3*a*B*e)*Sqrt[d + e*x])/(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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], 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) \text {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*}

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Mathematica [A]
time = 0.78, size = 143, normalized size = 0.91 \begin {gather*} \frac {\frac {\sqrt {b} \sqrt {d+e x} \left (A b (2 b d-a e+b e x)+B \left (-3 a^2 e+4 b^2 d x+a b (2 d-5 e x)\right )\right )}{(-b d+a e) (a+b x)^2}+\frac {e (-4 b B d+A b e+3 a B e) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{3/2}}}{4 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 152, normalized size = 0.97

method	result	size
derivativedivides	\(2 e \left (\frac {\frac {\left (A b e -5 B a e +4 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{8 \left (a e -b d \right ) b}-\frac {\left (A b e +3 B a e -4 B b d \right ) \sqrt {e x +d}}{8 b^{2}}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {\left (A b e +3 B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right ) b^{2} \sqrt {\left (a e -b d \right ) b}}\right )\)	\(152\)
default	\(2 e \left (\frac {\frac {\left (A b e -5 B a e +4 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{8 \left (a e -b d \right ) b}-\frac {\left (A b e +3 B a e -4 B b d \right ) \sqrt {e x +d}}{8 b^{2}}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {\left (A b e +3 B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right ) b^{2} \sqrt {\left (a e -b d \right ) b}}\right )\)	\(152\)

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,method=_RETURNVERBOSE)

[Out]

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

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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)*(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (149) = 298\).
time = 0.83, size = 705, normalized size = 4.49 \begin {gather*} \left [-\frac {\sqrt {b^{2} d - a b e} {\left ({\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} e^{2} - 4 \, {\left (B b^{3} d x^{2} + 2 \, B a b^{2} d x + B a^{2} b d\right )} e\right )} \log \left (\frac {2 \, b d + {\left (b x - a\right )} e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {x e + d}}{b x + a}\right ) + 2 \, {\left (4 \, B b^{4} d^{2} x + 2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} + {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} e^{2} - {\left ({\left (9 \, B a b^{3} - A b^{4}\right )} d x + {\left (5 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} d\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{7} d^{2} x^{2} + 2 \, a b^{6} d^{2} x + a^{2} b^{5} d^{2} + {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )} e^{2} - 2 \, {\left (a b^{6} d x^{2} + 2 \, a^{2} b^{5} d x + a^{3} b^{4} d\right )} e\right )}}, -\frac {\sqrt {-b^{2} d + a b e} {\left ({\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} e^{2} - 4 \, {\left (B b^{3} d x^{2} + 2 \, B a b^{2} d x + B a^{2} b d\right )} e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {x e + d}}{b x e + b d}\right ) + {\left (4 \, B b^{4} d^{2} x + 2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} + {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} e^{2} - {\left ({\left (9 \, B a b^{3} - A b^{4}\right )} d x + {\left (5 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} d\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{7} d^{2} x^{2} + 2 \, a b^{6} d^{2} x + a^{2} b^{5} d^{2} + {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )} e^{2} - 2 \, {\left (a b^{6} d x^{2} + 2 \, a^{2} b^{5} d x + a^{3} b^{4} d\right )} e\right )}}\right ] \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Giac [A]
time = 1.18, size = 245, normalized size = 1.56 \begin {gather*} \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}} \end {gather*}

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)

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Mupad [B]
time = 0.22, size = 222, normalized size = 1.41 \begin {gather*} \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} \end {gather*}

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)

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