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

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

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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} \begin {gather*} \frac {(b d-a e) (a B e-4 A b e+3 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 b^{3/2} e^{5/2}}-\frac {\sqrt {a+b x} \sqrt {d+e x} (a B e-4 A b e+3 b B d)}{4 b e^2}+\frac {B (a+b x)^{3/2} \sqrt {d+e x}}{2 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.76, size = 185, normalized size = 1.32 \begin {gather*} \frac {\sqrt {d+e x} \left (2 B e^2 (a+b x)^2-\frac {(a B e-4 A b e+3 b B d) \left (e (a+b x) \sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}}-\sqrt {e} \sqrt {a+b x} (b d-a e) \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )\right )}{\sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{4 b e^3 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.61, size = 194, normalized size = 1.39 \begin {gather*} \frac {\sqrt {\frac {b}{e}} \left (a^2 B e^2-4 a A b e^2+2 a b B d e+4 A b^2 d e-3 b^2 B d^2\right ) \log \left (\sqrt {a+\frac {b (d+e x)}{e}-\frac {b d}{e}}-\sqrt {\frac {b}{e}} \sqrt {d+e x}\right )}{4 b^2 e^2}+\frac {\sqrt {a+\frac {b (d+e x)}{e}-\frac {b d}{e}} \left (a B e \sqrt {d+e x}+4 A b e \sqrt {d+e x}+2 b B (d+e x)^{3/2}-5 b B d \sqrt {d+e x}\right )}{4 b e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.33, size = 364, normalized size = 2.60 \begin {gather*} \left [\frac {{\left (3 \, B b^{2} d^{2} - 2 \, {\left (B a b + 2 \, A b^{2}\right )} d e - {\left (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 - 3 \, B b^{2} d e + {\left (B a b + 4 \, A b^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{16 \, b^{2} e^{3}}, -\frac {{\left (3 \, B b^{2} d^{2} - 2 \, {\left (B a b + 2 \, A b^{2}\right )} d e - {\left (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 - 3 \, B b^{2} d e + {\left (B a b + 4 \, A b^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{8 \, b^{2} e^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.41, size = 175, normalized size = 1.25 \begin {gather*} \frac {{\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B e^{\left (-1\right )}}{b^{2}} - \frac {{\left (3 \, B b^{3} d e + B a b^{2} e^{2} - 4 \, A b^{3} e^{2}\right )} e^{\left (-3\right )}}{b^{4}}\right )} - \frac {{\left (3 \, B b^{2} d^{2} - 2 \, B a b d e - 4 \, A b^{2} d e - B a^{2} e^{2} + 4 \, A a b e^{2}\right )} e^{\left (-\frac {5}{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 )}{b^{\frac {3}{2}}}\right )} b}{4 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 376, normalized size = 2.69 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {e x +d}\, \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 )-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 )+3 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 +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a e -6 \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 \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 22.84, size = 872, normalized size = 6.23 \begin {gather*} \frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {B\,a^2\,b^2\,e^2}{2}+B\,a\,b^3\,d\,e-\frac {3\,B\,b^4\,d^2}{2}\right )}{e^6\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {7\,B\,a^2\,b\,e^2}{2}+23\,B\,a\,b^2\,d\,e+\frac {11\,B\,b^3\,d^2}{2}\right )}{e^5\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {7\,B\,a^2\,e^2}{2}+23\,B\,a\,b\,d\,e+\frac {11\,B\,b^2\,d^2}{2}\right )}{e^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {B\,a^2\,e^2}{2}+B\,a\,b\,d\,e-\frac {3\,B\,b^2\,d^2}{2}\right )}{b\,e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}-\frac {\sqrt {a}\,\sqrt {d}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (32\,B\,d\,b^2+16\,B\,a\,e\,b\right )}{e^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}-\frac {8\,B\,a^{3/2}\,\sqrt {d}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}-\frac {8\,B\,a^{3/2}\,b^2\,\sqrt {d}\,{\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 (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (2\,A\,d\,b^2+2\,A\,a\,e\,b\right )}{e^3\,\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}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}-\frac {8\,A\,\sqrt {a}\,b\,\sqrt {d}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e^2\,{\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 )}{\sqrt {b}\,e^{3/2}}-\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 (a\,e+3\,b\,d\right )}{2\,b^{3/2}\,e^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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