3.18.62 \(\int \frac {A+B x}{(a+b x)^3 \sqrt {d+e x}} \, dx\) [1762]

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

[Out]

1/4*e*(-3*A*b*e-B*a*e+4*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(3/2)/(-a*e+b*d)^(5/2)-1/2*(A
*b-B*a)*(e*x+d)^(1/2)/b/(-a*e+b*d)/(b*x+a)^2-1/4*(-3*A*b*e-B*a*e+4*B*b*d)*(e*x+d)^(1/2)/b/(-a*e+b*d)^2/(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, 44, 65, 214} \begin {gather*} \frac {e (-a B e-3 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^{3/2} (b d-a e)^{5/2}}-\frac {\sqrt {d+e x} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}-\frac {\sqrt {d+e x} (-a B e-3 A b e+4 b B d)}{4 b (a+b x) (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((A*b - a*B)*Sqrt[d + e*x])/(b*(b*d - a*e)*(a + b*x)^2) - ((4*b*B*d - 3*A*b*e - a*B*e)*Sqrt[d + e*x])/(4*
b*(b*d - a*e)^2*(a + b*x)) + (e*(4*b*B*d - 3*A*b*e - a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/
(4*b^(3/2)*(b*d - a*e)^(5/2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[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}{(a+b x)^3 \sqrt {d+e x}} \, dx &=-\frac {(A b-a B) \sqrt {d+e x}}{2 b (b d-a e) (a+b x)^2}+\frac {(4 b B d-3 A b e-a B e) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{4 b (b d-a e)}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{2 b (b d-a e) (a+b x)^2}-\frac {(4 b B d-3 A b e-a B e) \sqrt {d+e x}}{4 b (b d-a e)^2 (a+b x)}-\frac {(e (4 b B d-3 A b e-a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{2 b (b d-a e) (a+b x)^2}-\frac {(4 b B d-3 A b e-a B e) \sqrt {d+e x}}{4 b (b d-a e)^2 (a+b x)}-\frac {(4 b B d-3 A b e-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 (b d-a e)^2}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{2 b (b d-a e) (a+b x)^2}-\frac {(4 b B d-3 A b e-a B e) \sqrt {d+e x}}{4 b (b d-a e)^2 (a+b x)}+\frac {e (4 b B d-3 A b e-a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.78, size = 144, normalized size = 0.92 \begin {gather*} \frac {\frac {\sqrt {b} \sqrt {d+e x} \left (A b (-2 b d+5 a e+3 b e x)-B \left (a^2 e+4 b^2 d x+a b (2 d-e x)\right )\right )}{(b d-a e)^2 (a+b x)^2}+\frac {e (-4 b B d+3 A b e+a B e) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{5/2}}}{4 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 186, normalized size = 1.18

method	result	size
derivativedivides	\(2 e \left (\frac {\frac {\left (3 A b e +B a e -4 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{8 a^{2} e^{2}-16 a e b d +8 b^{2} d^{2}}+\frac {\left (5 A b e -B a e -4 B b d \right ) \sqrt {e x +d}}{8 \left (a e -b d \right ) b}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {\left (3 A b e +B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{2} e^{2}-2 a e b d +b^{2} d^{2}\right ) b \sqrt {\left (a e -b d \right ) b}}\right )\)	\(186\)
default	\(2 e \left (\frac {\frac {\left (3 A b e +B a e -4 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{8 a^{2} e^{2}-16 a e b d +8 b^{2} d^{2}}+\frac {\left (5 A b e -B a e -4 B b d \right ) \sqrt {e x +d}}{8 \left (a e -b d \right ) b}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {\left (3 A b e +B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{2} e^{2}-2 a e b d +b^{2} d^{2}\right ) b \sqrt {\left (a e -b d \right ) b}}\right )\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*((1/8*(3*A*b*e+B*a*e-4*B*b*d)/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)+1/8*(5*A*b*e-B*a*e-4*B*b*d)/(a*e-b
*d)/b*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+1/8*(3*A*b*e+B*a*e-4*B*b*d)/(a^2*e^2-2*a*b*d*e+b^2*d^2)/b/((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)/(b*x+a)^3/(e*x+d)^(1/2),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(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. 387 vs. \(2 (149) = 298\).
time = 0.92, size = 788, normalized size = 5.02 \begin {gather*} \left [\frac {\sqrt {b^{2} d - a b e} {\left ({\left (B a^{3} + 3 \, A a^{2} b + {\left (B a b^{2} + 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (B a^{2} b + 3 \, 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 (B a^{3} b - 5 \, A a^{2} b^{2} - {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x\right )} e^{2} - {\left ({\left (5 \, B a b^{3} + 3 \, A b^{4}\right )} d x + {\left (B a^{2} b^{2} + 7 \, A a b^{3}\right )} d\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{7} d^{3} x^{2} + 2 \, a b^{6} d^{3} x + a^{2} b^{5} d^{3} - {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )} e^{3} + 3 \, {\left (a^{2} b^{5} d x^{2} + 2 \, a^{3} b^{4} d x + a^{4} b^{3} d\right )} e^{2} - 3 \, {\left (a b^{6} d^{2} x^{2} + 2 \, a^{2} b^{5} d^{2} x + a^{3} b^{4} d^{2}\right )} e\right )}}, \frac {\sqrt {-b^{2} d + a b e} {\left ({\left (B a^{3} + 3 \, A a^{2} b + {\left (B a b^{2} + 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (B a^{2} b + 3 \, 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 (B a^{3} b - 5 \, A a^{2} b^{2} - {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x\right )} e^{2} - {\left ({\left (5 \, B a b^{3} + 3 \, A b^{4}\right )} d x + {\left (B a^{2} b^{2} + 7 \, A a b^{3}\right )} d\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{7} d^{3} x^{2} + 2 \, a b^{6} d^{3} x + a^{2} b^{5} d^{3} - {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )} e^{3} + 3 \, {\left (a^{2} b^{5} d x^{2} + 2 \, a^{3} b^{4} d x + a^{4} b^{3} d\right )} e^{2} - 3 \, {\left (a b^{6} d^{2} x^{2} + 2 \, a^{2} b^{5} d^{2} x + a^{3} b^{4} d^{2}\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]
time = 1.24, size = 266, normalized size = 1.69 \begin {gather*} -\frac {{\left (4 \, B b d e - B a e^{2} - 3 \, 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^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\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 - {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{2} - 3 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{2} + 3 \, \sqrt {x e + d} B a b d e^{2} + 5 \, \sqrt {x e + d} A b^{2} d e^{2} + \sqrt {x e + d} B a^{2} e^{3} - 5 \, \sqrt {x e + d} A a b e^{3}}{4 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\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)/(b*x+a)^3/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.34, size = 228, normalized size = 1.45 \begin {gather*} \frac {\frac {{\left (d+e\,x\right )}^{3/2}\,\left (3\,A\,b\,e^2+B\,a\,e^2-4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^2}-\frac {\sqrt {d+e\,x}\,\left (B\,a\,e^2-5\,A\,b\,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}+\frac {e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (3\,A\,b\,e+B\,a\,e-4\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (3\,A\,b\,e^2+B\,a\,e^2-4\,B\,b\,d\,e\right )}\right )\,\left (3\,A\,b\,e+B\,a\,e-4\,B\,b\,d\right )}{4\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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