3.18.55 \(\int \frac {A+B x}{(a+b x)^2 (d+e x)^{3/2}} \, dx\) [1755]

Optimal. Leaf size=140 \[ \frac {2 b B d-3 A b e+a B e}{b (b d-a e)^2 \sqrt {d+e x}}-\frac {A b-a B}{b (b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(2 b B d-3 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 140, 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, 53, 65, 214} \begin {gather*} -\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}+\frac {a B e-3 A b e+2 b B d}{b \sqrt {d+e x} (b d-a e)^2}-\frac {(a B e-3 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)^2*(d + e*x)^(3/2)),x]

[Out]

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

Rule 53

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

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Mathematica [A]
time = 0.51, size = 122, normalized size = 0.87 \begin {gather*} \frac {B (3 a d+2 b d x+a e x)-A (2 a e+b (d+3 e x))}{(b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {(2 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 )}{\sqrt {b} (-b d+a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)^2*(d + e*x)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.10, size = 130, normalized size = 0.93

method	result	size
derivativedivides	\(-\frac {2 \left (A e -B d \right )}{\left (a e -b d \right )^{2} \sqrt {e x +d}}-\frac {2 \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (3 A b e -B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2}}\)	\(130\)
default	\(-\frac {2 \left (A e -B d \right )}{\left (a e -b d \right )^{2} \sqrt {e x +d}}-\frac {2 \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (3 A b e -B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(A*e-B*d)/(a*e-b*d)^2/(e*x+d)^(1/2)-2/(a*e-b*d)^2*((1/2*A*b*e-1/2*B*a*e)*(e*x+d)^(1/2)/(b*(e*x+d)+a*e-b*d)+
1/2*(3*A*b*e-B*a*e-2*B*b*d)/((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)^2/(e*x+d)^(3/2),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. 383 vs. \(2 (136) = 272\).
time = 0.76, size = 782, normalized size = 5.59 \begin {gather*} \left [-\frac {{\left (2 \, B b^{2} d^{2} x + 2 \, B a b d^{2} + {\left ({\left (B a b - 3 \, A b^{2}\right )} x^{2} + {\left (B a^{2} - 3 \, A a b\right )} x\right )} e^{2} + {\left (2 \, B b^{2} d x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d x + {\left (B a^{2} - 3 \, A a b\right )} d\right )} e\right )} \sqrt {b^{2} d - a b e} \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 (2 \, B b^{3} d^{2} x + {\left (3 \, B a b^{2} - A b^{3}\right )} d^{2} + {\left (2 \, A a^{2} b - {\left (B a^{2} b - 3 \, A a b^{2}\right )} x\right )} e^{2} - {\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} d x + {\left (3 \, B a^{2} b + A a b^{2}\right )} d\right )} e\right )} \sqrt {x e + d}}{2 \, {\left (b^{5} d^{4} x + a b^{4} d^{4} - {\left (a^{3} b^{2} x^{2} + a^{4} b x\right )} e^{4} + {\left (3 \, a^{2} b^{3} d x^{2} + 2 \, a^{3} b^{2} d x - a^{4} b d\right )} e^{3} - 3 \, {\left (a b^{4} d^{2} x^{2} - a^{3} b^{2} d^{2}\right )} e^{2} + {\left (b^{5} d^{3} x^{2} - 2 \, a b^{4} d^{3} x - 3 \, a^{2} b^{3} d^{3}\right )} e\right )}}, \frac {{\left (2 \, B b^{2} d^{2} x + 2 \, B a b d^{2} + {\left ({\left (B a b - 3 \, A b^{2}\right )} x^{2} + {\left (B a^{2} - 3 \, A a b\right )} x\right )} e^{2} + {\left (2 \, B b^{2} d x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d x + {\left (B a^{2} - 3 \, A a b\right )} d\right )} e\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {x e + d}}{b x e + b d}\right ) + {\left (2 \, B b^{3} d^{2} x + {\left (3 \, B a b^{2} - A b^{3}\right )} d^{2} + {\left (2 \, A a^{2} b - {\left (B a^{2} b - 3 \, A a b^{2}\right )} x\right )} e^{2} - {\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} d x + {\left (3 \, B a^{2} b + A a b^{2}\right )} d\right )} e\right )} \sqrt {x e + d}}{b^{5} d^{4} x + a b^{4} d^{4} - {\left (a^{3} b^{2} x^{2} + a^{4} b x\right )} e^{4} + {\left (3 \, a^{2} b^{3} d x^{2} + 2 \, a^{3} b^{2} d x - a^{4} b d\right )} e^{3} - 3 \, {\left (a b^{4} d^{2} x^{2} - a^{3} b^{2} d^{2}\right )} e^{2} + {\left (b^{5} d^{3} x^{2} - 2 \, a b^{4} d^{3} x - 3 \, a^{2} b^{3} d^{3}\right )} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]
time = 0.99, size = 204, normalized size = 1.46 \begin {gather*} \frac {{\left (2 \, B b d + B a e - 3 \, A b e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, {\left (x e + d\right )} B b d - 2 \, B b d^{2} + {\left (x e + d\right )} B a e - 3 \, {\left (x e + d\right )} A b e + 2 \, B a d e + 2 \, A b d e - 2 \, A a e^{2}}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left ({\left (x e + d\right )}^{\frac {3}{2}} b - \sqrt {x e + d} b d + \sqrt {x e + d} a e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.20, size = 156, normalized size = 1.11 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^{5/2}}\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{\sqrt {b}\,{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,e-B\,d\right )}{a\,e-b\,d}-\frac {\left (d+e\,x\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^2}}{b\,{\left (d+e\,x\right )}^{3/2}+\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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