3.18.63 \(\int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx\) [1763]

Optimal. Leaf size=197 \[ -\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {3 e (4 b B d-5 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 44, 53, 65, 214} \begin {gather*} -\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}-\frac {3 e (a B e-5 A b e+4 b B d)}{4 b \sqrt {d+e x} (b d-a e)^3}-\frac {a B e-5 A b e+4 b B d}{4 b (a+b x) \sqrt {d+e x} (b d-a e)^2}+\frac {3 e (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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)^3 (d+e x)^{3/2}} \, dx &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {(4 b B d-5 A b e+a B e) \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{4 b (b d-a e)}\\ &=-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {(3 e (4 b B d-5 A b e+a B e)) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {(3 e (4 b B d-5 A b e+a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^3}\\ &=-\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {(3 (4 b B d-5 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 d-a e)^3}\\ &=-\frac {3 e (4 b B d-5 A b e+a B e)}{4 b (b d-a e)^3 \sqrt {d+e x}}-\frac {A b-a B}{2 b (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {3 e (4 b B d-5 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 1.19, size = 194, normalized size = 0.98 \begin {gather*} \frac {1}{4} \left (\frac {-B \left (4 b^2 d x (d+3 e x)+a^2 e (13 d+5 e x)+a b \left (2 d^2+21 d e x+3 e^2 x^2\right )\right )+A \left (8 a^2 e^2+a b e (9 d+25 e x)+b^2 \left (-2 d^2+5 d e x+15 e^2 x^2\right )\right )}{(b d-a e)^3 (a+b x)^2 \sqrt {d+e x}}+\frac {3 e (4 b B d-5 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)^{7/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(B*(4*b^2*d*x*(d + 3*e*x) + a^2*e*(13*d + 5*e*x) + a*b*(2*d^2 + 21*d*e*x + 3*e^2*x^2))) + A*(8*a^2*e^2 + a*
b*e*(9*d + 25*e*x) + b^2*(-2*d^2 + 5*d*e*x + 15*e^2*x^2)))/((b*d - a*e)^3*(a + b*x)^2*Sqrt[d + e*x]) + (3*e*(4
*b*B*d - 5*A*b*e + a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(Sqrt[b]*(-(b*d) + a*e)^(7/2)))/
4

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Maple [A]
time = 0.09, size = 195, normalized size = 0.99

method	result	size
derivativedivides	\(2 e \left (-\frac {A e -B d}{\left (a e -b d \right )^{3} \sqrt {e x +d}}-\frac {\frac {\left (\frac {7}{8} A \,b^{2} e -\frac {3}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (\frac {9}{8} A a b \,e^{2}-\frac {9}{8} A \,b^{2} d e -\frac {5}{8} B \,a^{2} e^{2}+\frac {1}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {3 \left (5 A b e -B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{\left (a e -b d \right )^{3}}\right )\)	\(195\)
default	\(2 e \left (-\frac {A e -B d}{\left (a e -b d \right )^{3} \sqrt {e x +d}}-\frac {\frac {\left (\frac {7}{8} A \,b^{2} e -\frac {3}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (\frac {9}{8} A a b \,e^{2}-\frac {9}{8} A \,b^{2} d e -\frac {5}{8} B \,a^{2} e^{2}+\frac {1}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {3 \left (5 A b e -B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{\left (a e -b d \right )^{3}}\right )\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*(-(A*e-B*d)/(a*e-b*d)^3/(e*x+d)^(1/2)-1/(a*e-b*d)^3*(((7/8*A*b^2*e-3/8*B*a*b*e-1/2*b^2*B*d)*(e*x+d)^(3/2)+
(9/8*A*a*b*e^2-9/8*A*b^2*d*e-5/8*B*a^2*e^2+1/8*B*a*b*d*e+1/2*b^2*B*d^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+3
/8*(5*A*b*e-B*a*e-4*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)^3/(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. 690 vs. \(2 (190) = 380\).
time = 1.25, size = 1395, normalized size = 7.08 \begin {gather*} \left [\frac {3 \, \sqrt {b^{2} d - a b e} {\left ({\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} e^{3} + {\left (4 \, B b^{3} d x^{3} + {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d x^{2} + 2 \, {\left (3 \, B a^{2} b - 5 \, A a b^{2}\right )} d x + {\left (B a^{3} - 5 \, A a^{2} b\right )} d\right )} e^{2} + 4 \, {\left (B b^{3} d^{2} x^{2} + 2 \, B a b^{2} d^{2} x + B a^{2} b d^{2}\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^{3} x + 2 \, {\left (B a b^{3} + A b^{4}\right )} d^{3} + {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} e^{3} - {\left (3 \, {\left (3 \, B a b^{3} + 5 \, A b^{4}\right )} d x^{2} + 4 \, {\left (4 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} d x + {\left (13 \, B a^{3} b - A a^{2} b^{2}\right )} d\right )} e^{2} + {\left (12 \, B b^{4} d^{2} x^{2} + {\left (17 \, B a b^{3} - 5 \, A b^{4}\right )} d^{2} x + 11 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d^{2}\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{7} d^{5} x^{2} + 2 \, a b^{6} d^{5} x + a^{2} b^{5} d^{5} + {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )} e^{5} - {\left (4 \, a^{3} b^{4} d x^{3} + 7 \, a^{4} b^{3} d x^{2} + 2 \, a^{5} b^{2} d x - a^{6} b d\right )} e^{4} + 2 \, {\left (3 \, a^{2} b^{5} d^{2} x^{3} + 4 \, a^{3} b^{4} d^{2} x^{2} - a^{4} b^{3} d^{2} x - 2 \, a^{5} b^{2} d^{2}\right )} e^{3} - 2 \, {\left (2 \, a b^{6} d^{3} x^{3} + a^{2} b^{5} d^{3} x^{2} - 4 \, a^{3} b^{4} d^{3} x - 3 \, a^{4} b^{3} d^{3}\right )} e^{2} + {\left (b^{7} d^{4} x^{3} - 2 \, a b^{6} d^{4} x^{2} - 7 \, a^{2} b^{5} d^{4} x - 4 \, a^{3} b^{4} d^{4}\right )} e\right )}}, -\frac {3 \, \sqrt {-b^{2} d + a b e} {\left ({\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} e^{3} + {\left (4 \, B b^{3} d x^{3} + {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d x^{2} + 2 \, {\left (3 \, B a^{2} b - 5 \, A a b^{2}\right )} d x + {\left (B a^{3} - 5 \, A a^{2} b\right )} d\right )} e^{2} + 4 \, {\left (B b^{3} d^{2} x^{2} + 2 \, B a b^{2} d^{2} x + B a^{2} b d^{2}\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^{3} x + 2 \, {\left (B a b^{3} + A b^{4}\right )} d^{3} + {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} e^{3} - {\left (3 \, {\left (3 \, B a b^{3} + 5 \, A b^{4}\right )} d x^{2} + 4 \, {\left (4 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} d x + {\left (13 \, B a^{3} b - A a^{2} b^{2}\right )} d\right )} e^{2} + {\left (12 \, B b^{4} d^{2} x^{2} + {\left (17 \, B a b^{3} - 5 \, A b^{4}\right )} d^{2} x + 11 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d^{2}\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{7} d^{5} x^{2} + 2 \, a b^{6} d^{5} x + a^{2} b^{5} d^{5} + {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )} e^{5} - {\left (4 \, a^{3} b^{4} d x^{3} + 7 \, a^{4} b^{3} d x^{2} + 2 \, a^{5} b^{2} d x - a^{6} b d\right )} e^{4} + 2 \, {\left (3 \, a^{2} b^{5} d^{2} x^{3} + 4 \, a^{3} b^{4} d^{2} x^{2} - a^{4} b^{3} d^{2} x - 2 \, a^{5} b^{2} d^{2}\right )} e^{3} - 2 \, {\left (2 \, a b^{6} d^{3} x^{3} + a^{2} b^{5} d^{3} x^{2} - 4 \, a^{3} b^{4} d^{3} x - 3 \, a^{4} b^{3} d^{3}\right )} e^{2} + {\left (b^{7} d^{4} x^{3} - 2 \, a b^{6} d^{4} x^{2} - 7 \, a^{2} b^{5} d^{4} x - 4 \, a^{3} b^{4} d^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*sqrt(b^2*d - a*b*e)*(((B*a*b^2 - 5*A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*e
^3 + (4*B*b^3*d*x^3 + (9*B*a*b^2 - 5*A*b^3)*d*x^2 + 2*(3*B*a^2*b - 5*A*a*b^2)*d*x + (B*a^3 - 5*A*a^2*b)*d)*e^2
+ 4*(B*b^3*d^2*x^2 + 2*B*a*b^2*d^2*x + B*a^2*b*d^2)*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^3*x + 2*(B*a*b^3 + A*b^4)*d^3 + (8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)*x^2
- 5*(B*a^3*b - 5*A*a^2*b^2)*x)*e^3 - (3*(3*B*a*b^3 + 5*A*b^4)*d*x^2 + 4*(4*B*a^2*b^2 + 5*A*a*b^3)*d*x + (13*B
*a^3*b - A*a^2*b^2)*d)*e^2 + (12*B*b^4*d^2*x^2 + (17*B*a*b^3 - 5*A*b^4)*d^2*x + 11*(B*a^2*b^2 - A*a*b^3)*d^2)*
e)*sqrt(x*e + d))/(b^7*d^5*x^2 + 2*a*b^6*d^5*x + a^2*b^5*d^5 + (a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x)*e^5 - (
4*a^3*b^4*d*x^3 + 7*a^4*b^3*d*x^2 + 2*a^5*b^2*d*x - a^6*b*d)*e^4 + 2*(3*a^2*b^5*d^2*x^3 + 4*a^3*b^4*d^2*x^2 -
a^4*b^3*d^2*x - 2*a^5*b^2*d^2)*e^3 - 2*(2*a*b^6*d^3*x^3 + a^2*b^5*d^3*x^2 - 4*a^3*b^4*d^3*x - 3*a^4*b^3*d^3)*e
^2 + (b^7*d^4*x^3 - 2*a*b^6*d^4*x^2 - 7*a^2*b^5*d^4*x - 4*a^3*b^4*d^4)*e), -1/4*(3*sqrt(-b^2*d + a*b*e)*(((B*a
*b^2 - 5*A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*e^3 + (4*B*b^3*d*x^3 + (9*B*a*b^2 -
5*A*b^3)*d*x^2 + 2*(3*B*a^2*b - 5*A*a*b^2)*d*x + (B*a^3 - 5*A*a^2*b)*d)*e^2 + 4*(B*b^3*d^2*x^2 + 2*B*a*b^2*d^
2*x + B*a^2*b*d^2)*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(x*e + d)/(b*x*e + b*d)) + (4*B*b^4*d^3*x + 2*(B*a*b^3 +
A*b^4)*d^3 + (8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)*x^2 - 5*(B*a^3*b - 5*A*a^2*b^2)*x)*e^3 - (3*(3*B*a*b^3 +
5*A*b^4)*d*x^2 + 4*(4*B*a^2*b^2 + 5*A*a*b^3)*d*x + (13*B*a^3*b - A*a^2*b^2)*d)*e^2 + (12*B*b^4*d^2*x^2 + (17*B
*a*b^3 - 5*A*b^4)*d^2*x + 11*(B*a^2*b^2 - A*a*b^3)*d^2)*e)*sqrt(x*e + d))/(b^7*d^5*x^2 + 2*a*b^6*d^5*x + a^2*b
^5*d^5 + (a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x)*e^5 - (4*a^3*b^4*d*x^3 + 7*a^4*b^3*d*x^2 + 2*a^5*b^2*d*x - a^
6*b*d)*e^4 + 2*(3*a^2*b^5*d^2*x^3 + 4*a^3*b^4*d^2*x^2 - a^4*b^3*d^2*x - 2*a^5*b^2*d^2)*e^3 - 2*(2*a*b^6*d^3*x^
3 + a^2*b^5*d^3*x^2 - 4*a^3*b^4*d^3*x - 3*a^4*b^3*d^3)*e^2 + (b^7*d^4*x^3 - 2*a*b^6*d^4*x^2 - 7*a^2*b^5*d^4*x
- 4*a^3*b^4*d^4)*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)**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.96, size = 346, normalized size = 1.76 \begin {gather*} -\frac {3 \, {\left (4 \, B b d e + B a e^{2} - 5 \, 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^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (B d e - A e^{2}\right )}}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {x e + d}} - \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 + 3 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{2} - 7 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{2} - \sqrt {x e + d} B a b d e^{2} + 9 \, \sqrt {x e + d} A b^{2} d e^{2} + 5 \, \sqrt {x e + d} B a^{2} e^{3} - 9 \, \sqrt {x e + d} A a b e^{3}}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\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)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.46, size = 296, normalized size = 1.50 \begin {gather*} \frac {\frac {5\,\left (d+e\,x\right )\,\left (B\,a\,e^2-5\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^2}-\frac {2\,\left (A\,e^2-B\,d\,e\right )}{a\,e-b\,d}+\frac {3\,b\,{\left (d+e\,x\right )}^2\,\left (B\,a\,e^2-5\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^3}}{b^2\,{\left (d+e\,x\right )}^{5/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{3/2}+\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}+\frac {3\,e\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (B\,a\,e-5\,A\,b\,e+4\,B\,b\,d\right )\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}\,\left (3\,B\,a\,e^2-15\,A\,b\,e^2+12\,B\,b\,d\,e\right )}\right )\,\left (B\,a\,e-5\,A\,b\,e+4\,B\,b\,d\right )}{4\,\sqrt {b}\,{\left (a\,e-b\,d\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*(d + e*x)*(B*a*e^2 - 5*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^2) - (2*(A*e^2 - B*d*e))/(a*e - b*d) + (3*b*(d
+ e*x)^2*(B*a*e^2 - 5*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^3))/(b^2*(d + e*x)^(5/2) - (2*b^2*d - 2*a*b*e)*(d
+ e*x)^(3/2) + (d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e)) + (3*e*atan((3*b^(1/2)*e*(d + e*x)^(1/2)*(B*a*
e - 5*A*b*e + 4*B*b*d)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/((a*e - b*d)^(7/2)*(3*B*a*e^2 - 15
*A*b*e^2 + 12*B*b*d*e)))*(B*a*e - 5*A*b*e + 4*B*b*d))/(4*b^(1/2)*(a*e - b*d)^(7/2))

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