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

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

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Rubi [A]  time = 0.12, 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, 51, 63, 208} \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]

-((A*b - a*B)*Sqrt[d + e*x])/(2*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 51

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 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}{(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) \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 (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 = 0.35, size = 149, normalized size = 0.95 \begin {gather*} \frac {\sqrt {d+e x} \left (\frac {e (-a B e-3 A b e+4 b B d) \left (\frac {a e-b d}{e (a+b x)}+\frac {\sqrt {a e-b d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )}{\sqrt {b} \sqrt {d+e x}}\right )}{2 (b d-a e)^2}+\frac {a B-A b}{(a+b x)^2}\right )}{2 b (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.74, size = 212, normalized size = 1.35 \begin {gather*} \frac {\left (-a B e^2-3 A b e^2+4 b B d e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{3/2} (b d-a e)^2 \sqrt {a e-b d}}-\frac {e \sqrt {d+e x} \left (a^2 B e^2-5 a A b e^2-a b B e (d+e x)+3 a b B d e-3 A b^2 e (d+e x)+5 A b^2 d e-4 b^2 B d^2+4 b^2 B d (d+e x)\right )}{4 b (b d-a e)^2 (-a e-b (d+e x)+b d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.03, size = 808, normalized size = 5.15 \begin {gather*} \left [-\frac {{\left (4 \, B a^{2} b d e - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (B a b^{2} + 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (B a^{2} b + 3 \, 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 (B a^{2} b^{2} + 7 \, A a b^{3}\right )} d e - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (5 \, B a b^{3} + 3 \, A b^{4}\right )} d e + {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{5} d^{3} - 3 \, a^{3} b^{4} d^{2} e + 3 \, a^{4} b^{3} d e^{2} - a^{5} b^{2} e^{3} + {\left (b^{7} d^{3} - 3 \, a b^{6} d^{2} e + 3 \, a^{2} b^{5} d e^{2} - a^{3} b^{4} e^{3}\right )} x^{2} + 2 \, {\left (a b^{6} d^{3} - 3 \, a^{2} b^{5} d^{2} e + 3 \, a^{3} b^{4} d e^{2} - a^{4} b^{3} e^{3}\right )} x\right )}}, -\frac {{\left (4 \, B a^{2} b d e - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (B a b^{2} + 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (B a^{2} b + 3 \, 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 (B a^{2} b^{2} + 7 \, A a b^{3}\right )} d e - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (5 \, B a b^{3} + 3 \, A b^{4}\right )} d e + {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{5} d^{3} - 3 \, a^{3} b^{4} d^{2} e + 3 \, a^{4} b^{3} d e^{2} - a^{5} b^{2} e^{3} + {\left (b^{7} d^{3} - 3 \, a b^{6} d^{2} e + 3 \, a^{2} b^{5} d e^{2} - a^{3} b^{4} e^{3}\right )} x^{2} + 2 \, {\left (a b^{6} d^{3} - 3 \, a^{2} b^{5} d^{2} e + 3 \, a^{3} b^{4} d e^{2} - a^{4} b^{3} e^{3}\right )} x\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*((4*B*a^2*b*d*e - (B*a^3 + 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (B*a*b^2 + 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d
*e - (B*a^2*b + 3*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 - (B*a^2*b^2 + 7*A*a*b^3)*d*e - (B*a^3*b - 5*A*a^2*b^2)*e^2 +
(4*B*b^4*d^2 - (5*B*a*b^3 + 3*A*b^4)*d*e + (B*a^2*b^2 + 3*A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^3 - 3*a^
3*b^4*d^2*e + 3*a^4*b^3*d*e^2 - a^5*b^2*e^3 + (b^7*d^3 - 3*a*b^6*d^2*e + 3*a^2*b^5*d*e^2 - a^3*b^4*e^3)*x^2 +
2*(a*b^6*d^3 - 3*a^2*b^5*d^2*e + 3*a^3*b^4*d*e^2 - a^4*b^3*e^3)*x), -1/4*((4*B*a^2*b*d*e - (B*a^3 + 3*A*a^2*b)
*e^2 + (4*B*b^3*d*e - (B*a*b^2 + 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (B*a^2*b + 3*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 - (B*a^2*b^2
+ 7*A*a*b^3)*d*e - (B*a^3*b - 5*A*a^2*b^2)*e^2 + (4*B*b^4*d^2 - (5*B*a*b^3 + 3*A*b^4)*d*e + (B*a^2*b^2 + 3*A*a
*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^3 - 3*a^3*b^4*d^2*e + 3*a^4*b^3*d*e^2 - a^5*b^2*e^3 + (b^7*d^3 - 3*a*b
^6*d^2*e + 3*a^2*b^5*d*e^2 - a^3*b^4*e^3)*x^2 + 2*(a*b^6*d^3 - 3*a^2*b^5*d^2*e + 3*a^3*b^4*d*e^2 - a^4*b^3*e^3
)*x)]

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giac [A]  time = 1.26, 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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maple [B]  time = 0.02, size = 436, normalized size = 2.78 \begin {gather*} \frac {3 \left (e x +d \right )^{\frac {3}{2}} A b \,e^{2}}{4 \left (b x e +a e \right )^{2} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {3 A \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}}+\frac {B a \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} B a \,e^{2}}{4 \left (b x e +a e \right )^{2} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {\left (e x +d \right )^{\frac {3}{2}} B b d e}{\left (b x e +a e \right )^{2} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {B d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}}+\frac {5 \sqrt {e x +d}\, A \,e^{2}}{4 \left (b x e +a e \right )^{2} \left (a e -b d \right )}-\frac {\sqrt {e x +d}\, B a \,e^{2}}{4 \left (b x e +a e \right )^{2} \left (a e -b d \right ) b}-\frac {\sqrt {e x +d}\, B d e}{\left (b x e +a e \right )^{2} \left (a e -b d \right )} \end {gather*}

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)

[Out]

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

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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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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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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sympy [F(-1)]  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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