3.2238 \(\int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a+b x}} \, dx\)

Optimal. Leaf size=140 \[ -\frac {(b d-a e) (3 a B e-4 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 b^{5/2} e^{3/2}}-\frac {\sqrt {a+b x} \sqrt {d+e x} (3 a B e-4 A b e+b B d)}{4 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{3/2}}{2 b e} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {80, 50, 63, 217, 206} \[ -\frac {(b d-a e) (3 a B e-4 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 b^{5/2} e^{3/2}}-\frac {\sqrt {a+b x} \sqrt {d+e x} (3 a B e-4 A b e+b B d)}{4 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{3/2}}{2 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 135, normalized size = 0.96 \[ \frac {\sqrt {d+e x} \left (\sqrt {e} \sqrt {a+b x} (-3 a B e+4 A b e+b B (d+2 e x))-\frac {\sqrt {b d-a e} (3 a B e-4 A b e+b B d) \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{\sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{4 b^2 e^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.72, size = 366, normalized size = 2.61 \[ \left [\frac {{\left (B b^{2} d^{2} + 2 \, {\left (B a b - 2 \, A b^{2}\right )} d e - {\left (3 \, B a^{2} - 4 \, A a b\right )} e^{2}\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} - 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (2 \, B b^{2} e^{2} x + B b^{2} d e - {\left (3 \, B a b - 4 \, A b^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{16 \, b^{3} e^{2}}, \frac {{\left (B b^{2} d^{2} + 2 \, {\left (B a b - 2 \, A b^{2}\right )} d e - {\left (3 \, B a^{2} - 4 \, A a b\right )} e^{2}\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, B b^{2} e^{2} x + B b^{2} d e - {\left (3 \, B a b - 4 \, A b^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{8 \, b^{3} e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((B*b^2*d^2 + 2*(B*a*b - 2*A*b^2)*d*e - (3*B*a^2 - 4*A*a*b)*e^2)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 +
6*a*b*d*e + a^2*e^2 - 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x
) + 4*(2*B*b^2*e^2*x + B*b^2*d*e - (3*B*a*b - 4*A*b^2)*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^2), 1/8*((B*b^
2*d^2 + 2*(B*a*b - 2*A*b^2)*d*e - (3*B*a^2 - 4*A*a*b)*e^2)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b
*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) + 2*(2*B*b^2*e^2*x + B*b^2*d*
e - (3*B*a*b - 4*A*b^2)*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^2)]

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giac [B]  time = 1.14, size = 238, normalized size = 1.70 \[ -\frac {\frac {4 \, {\left (\frac {{\left (b^{2} d - a b e\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a}\right )} A {\left | b \right |}}{b^{2}} - \frac {{\left (\frac {{\left (b^{3} d^{2} + 2 \, a b^{2} d e - 3 \, a^{2} b e^{2}\right )} e^{\left (-\frac {3}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} {\left (2 \, b x + {\left (b d e - 5 \, a e^{2}\right )} e^{\left (-2\right )} + 2 \, a\right )} \sqrt {b x + a}\right )} B {\left | b \right |}}{b^{3}}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 375, normalized size = 2.68 \[ -\frac {\sqrt {e x +d}\, \sqrt {b x +a}\, \left (4 A a b \,e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-4 A \,b^{2} d e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-3 B \,a^{2} e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+2 B a b d e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+B \,b^{2} d^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b e x -8 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, A b e +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a e -2 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B b d \right )}{8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(4*A*a*b*e^2*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(
b*e)^(1/2))-4*A*b^2*d*e*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-3*B*a^2*e^
2*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+2*B*a*b*d*e*ln(1/2*(2*b*e*x+a*e+
b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+B*b^2*d^2*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^
(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-4*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b*e*x-8*(b*e)^(1/2)*((b*x+a)*(e*x+d))^
(1/2)*A*b*e+6*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*e-2*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*B*b*d)/((b*x+a)*
(e*x+d))^(1/2)/b^2/e/(b*e)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(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 zero or nonzero?

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mupad [B]  time = 22.14, size = 866, normalized size = 6.19 \[ \frac {\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {11\,B\,a^2\,e^2}{2}+23\,B\,a\,b\,d\,e+\frac {7\,B\,b^2\,d^2}{2}\right )}{e^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}+\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (-\frac {3\,B\,a^2\,b\,e^2}{2}+B\,a\,b^2\,d\,e+\frac {B\,b^3\,d^2}{2}\right )}{e^5\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (-\frac {3\,B\,a^2\,e^2}{2}+B\,a\,b\,d\,e+\frac {B\,b^2\,d^2}{2}\right )}{b^2\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {11\,B\,a^2\,e^2}{2}+23\,B\,a\,b\,d\,e+\frac {7\,B\,b^2\,d^2}{2}\right )}{b\,e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}-\frac {\sqrt {a}\,\sqrt {d}\,\left (32\,B\,a\,e+16\,B\,b\,d\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}-\frac {8\,B\,\sqrt {a}\,d^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}-\frac {8\,B\,\sqrt {a}\,b^2\,d^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^8}+\frac {b^4}{e^4}-\frac {4\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}-\frac {4\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}}+\frac {\frac {\left (2\,A\,a\,e+2\,A\,b\,d\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{e^2\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}+\frac {\left (2\,A\,a\,e+2\,A\,b\,d\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{b\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}-\frac {8\,A\,\sqrt {a}\,\sqrt {d}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}+\frac {b^2}{e^2}-\frac {2\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {\sqrt {e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )\,\left (a\,e-b\,d\right )}{b^{3/2}\,\sqrt {e}}+\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )\,\left (a\,e-b\,d\right )\,\left (3\,a\,e+b\,d\right )}{2\,b^{5/2}\,e^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x)^(1/2) - a^(1/2))^3*((11*B*a^2*e^2)/2 + (7*B*b^2*d^2)/2 + 23*B*a*b*d*e))/(e^4*((d + e*x)^(1/2) - d
^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))*((B*b^3*d^2)/2 - (3*B*a^2*b*e^2)/2 + B*a*b^2*d*e))/(e^5*((d + e*x)^(
1/2) - d^(1/2))) + (((a + b*x)^(1/2) - a^(1/2))^7*((B*b^2*d^2)/2 - (3*B*a^2*e^2)/2 + B*a*b*d*e))/(b^2*e^2*((d
+ e*x)^(1/2) - d^(1/2))^7) + (((a + b*x)^(1/2) - a^(1/2))^5*((11*B*a^2*e^2)/2 + (7*B*b^2*d^2)/2 + 23*B*a*b*d*e
))/(b*e^3*((d + e*x)^(1/2) - d^(1/2))^5) - (a^(1/2)*d^(1/2)*(32*B*a*e + 16*B*b*d)*((a + b*x)^(1/2) - a^(1/2))^
4)/(e^3*((d + e*x)^(1/2) - d^(1/2))^4) - (8*B*a^(1/2)*d^(3/2)*((a + b*x)^(1/2) - a^(1/2))^6)/(e^2*((d + e*x)^(
1/2) - d^(1/2))^6) - (8*B*a^(1/2)*b^2*d^(3/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(e^4*((d + e*x)^(1/2) - d^(1/2))^
2))/(((a + b*x)^(1/2) - a^(1/2))^8/((d + e*x)^(1/2) - d^(1/2))^8 + b^4/e^4 - (4*b^3*((a + b*x)^(1/2) - a^(1/2)
)^2)/(e^3*((d + e*x)^(1/2) - d^(1/2))^2) + (6*b^2*((a + b*x)^(1/2) - a^(1/2))^4)/(e^2*((d + e*x)^(1/2) - d^(1/
2))^4) - (4*b*((a + b*x)^(1/2) - a^(1/2))^6)/(e*((d + e*x)^(1/2) - d^(1/2))^6)) + (((2*A*a*e + 2*A*b*d)*((a +
b*x)^(1/2) - a^(1/2)))/(e^2*((d + e*x)^(1/2) - d^(1/2))) + ((2*A*a*e + 2*A*b*d)*((a + b*x)^(1/2) - a^(1/2))^3)
/(b*e*((d + e*x)^(1/2) - d^(1/2))^3) - (8*A*a^(1/2)*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(e*((d + e*x)^(1/2)
- d^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2))^4/((d + e*x)^(1/2) - d^(1/2))^4 + b^2/e^2 - (2*b*((a + b*x)^(1/2)
- a^(1/2))^2)/(e*((d + e*x)^(1/2) - d^(1/2))^2)) - (2*A*atanh((e^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*
((d + e*x)^(1/2) - d^(1/2))))*(a*e - b*d))/(b^(3/2)*e^(1/2)) + (B*atanh((e^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/
(b^(1/2)*((d + e*x)^(1/2) - d^(1/2))))*(a*e - b*d)*(3*a*e + b*d))/(2*b^(5/2)*e^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(1/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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