3.6.88 \(\int \frac {15 d^2+20 d e x+8 e^2 x^2}{\sqrt {a+b x} \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=122 \[ \frac {2 \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2} \sqrt {e}}+\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {2 \sqrt {a+b x} \sqrt {d+e x} (7 b d-5 a e)}{b^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {951, 80, 63, 217, 206} \begin {gather*} \frac {2 \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2} \sqrt {e}}+\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {2 \sqrt {a+b x} \sqrt {d+e x} (7 b d-5 a e)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(2*(7*b*d - 5*a*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/b^2 + (4*e*(a + b*x)^(3/2)*Sqrt[d + e*x])/b^2 + (2*(8*b^2*d^2
- 8*a*b*d*e + 3*a^2*e^2)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(5/2)*Sqrt[e])

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]

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {15 d^2+20 d e x+8 e^2 x^2}{\sqrt {a+b x} \sqrt {d+e x}} \, dx &=\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {\int \frac {2 e \left (15 b^2 d^2-6 a b d e-2 a^2 e^2\right )+4 b e^2 (7 b d-5 a e) x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{2 b^2 e}\\ &=\frac {2 (7 b d-5 a e) \sqrt {a+b x} \sqrt {d+e x}}{b^2}+\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {\left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{b^2}\\ &=\frac {2 (7 b d-5 a e) \sqrt {a+b x} \sqrt {d+e x}}{b^2}+\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {\left (2 \left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d-\frac {a e}{b}+\frac {e x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{b^3}\\ &=\frac {2 (7 b d-5 a e) \sqrt {a+b x} \sqrt {d+e x}}{b^2}+\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {\left (2 \left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {e x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {d+e x}}\right )}{b^3}\\ &=\frac {2 (7 b d-5 a e) \sqrt {a+b x} \sqrt {d+e x}}{b^2}+\frac {4 e (a+b x)^{3/2} \sqrt {d+e x}}{b^2}+\frac {2 \left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2} \sqrt {e}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.41, size = 135, normalized size = 1.11 \begin {gather*} \frac {2 \left (\frac {\sqrt {b d-a e} \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \sqrt {\frac {b (d+e x)}{b d-a e}} \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{\sqrt {e}}+b \sqrt {a+b x} (d+e x) (-3 a e+7 b d+2 b e x)\right )}{b^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(2*(b*Sqrt[a + b*x]*(d + e*x)*(7*b*d - 3*a*e + 2*b*e*x) + (Sqrt[b*d - a*e]*(8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)
*Sqrt[(b*(d + e*x))/(b*d - a*e)]*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/Sqrt[e]))/(b^3*Sqrt[d + e*x
])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.29, size = 196, normalized size = 1.61 \begin {gather*} \frac {2 \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{5/2} \sqrt {e}}+\frac {2 \sqrt {d+e x} \left (\frac {5 a^2 b e^2 (d+e x)}{a+b x}-3 a^2 e^3+\frac {7 b^3 d^2 (d+e x)}{a+b x}-\frac {12 a b^2 d e (d+e x)}{a+b x}+8 a b d e^2-5 b^2 d^2 e\right )}{b^2 \sqrt {a+b x} \left (\frac {b (d+e x)}{a+b x}-e\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(2*Sqrt[d + e*x]*(-5*b^2*d^2*e + 8*a*b*d*e^2 - 3*a^2*e^3 + (7*b^3*d^2*(d + e*x))/(a + b*x) - (12*a*b^2*d*e*(d
+ e*x))/(a + b*x) + (5*a^2*b*e^2*(d + e*x))/(a + b*x)))/(b^2*Sqrt[a + b*x]*(-e + (b*(d + e*x))/(a + b*x))^2) +
(2*(8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/(b^(5/2)*Sqr
t[e])

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 308, normalized size = 2.52 \begin {gather*} \left [\frac {{\left (8 \, b^{2} d^{2} - 8 \, a b d e + 3 \, a^{2} 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^{2} e^{2} x + 7 \, b^{2} d e - 3 \, a b e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{2 \, b^{3} e}, -\frac {{\left (8 \, b^{2} d^{2} - 8 \, a b d e + 3 \, a^{2} 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^{2} e^{2} x + 7 \, b^{2} d e - 3 \, a b e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{b^{3} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/2*((8*b^2*d^2 - 8*a*b*d*e + 3*a^2*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^2*e^2*x + 7*b^2*d*
e - 3*a*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e), -((8*b^2*d^2 - 8*a*b*d*e + 3*a^2*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^2*e^2*x + 7*b^2*d*e - 3*a*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e)]

________________________________________________________________________________________

giac [A]  time = 0.24, size = 145, normalized size = 1.19 \begin {gather*} \frac {2 \, {\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 )} e}{b^{3}} + \frac {{\left (7 \, b^{6} d e^{2} - 5 \, a b^{5} e^{3}\right )} e^{\left (-2\right )}}{b^{8}}\right )} - \frac {{\left (8 \, b^{2} d^{2} - 8 \, a b d e + 3 \, a^{2} e^{2}\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 )}{b^{\frac {5}{2}}}\right )} b}{{\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e/b^3 + (7*b^6*d*e^2 - 5*a*b^5*e^3)*e^(-2)/b
^8) - (8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)*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)))/b^(5/2))*b/abs(b)

________________________________________________________________________________________

maple [B]  time = 0.03, size = 247, normalized size = 2.02 \begin {gather*} \frac {\left (3 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 )-8 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 )+8 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 {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b e x -6 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a e +14 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b d \right ) \sqrt {e x +d}\, \sqrt {b x +a}}{\sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x)

[Out]

(3*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))*a^2*e^2-8*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))*a*b*d*e+8*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^2*d^2+4*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*x*b*e-6*(b*e)^(1/2)*((b*x+a)*(e*x
+d))^(1/2)*a*e+14*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b*d)*(e*x+d)^(1/2)*(b*x+a)^(1/2)/(b*e)^(1/2)/b^2/((b*x+a
)*(e*x+d))^(1/2)

________________________________________________________________________________________

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((8*e^2*x^2+20*d*e*x+15*d^2)/(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?

________________________________________________________________________________________

mupad [B]  time = 20.64, size = 893, normalized size = 7.32 \begin {gather*} \frac {\frac {\left (40\,b\,d^2+40\,a\,e\,d\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{e^2\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}-\frac {160\,\sqrt {a}\,d^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}+\frac {\left (40\,b\,d^2+40\,a\,e\,d\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{b\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}}{\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 {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (12\,a^2\,b\,e^2+8\,a\,b^2\,d\,e+12\,b^3\,d^2\right )}{e^4\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (44\,a^2\,e^2+200\,a\,b\,d\,e+44\,b^2\,d^2\right )}{e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (12\,a^2\,e^2+8\,a\,b\,d\,e+12\,b^2\,d^2\right )}{b^2\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (44\,a^2\,e^2+200\,a\,b\,d\,e+44\,b^2\,d^2\right )}{b\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}+\frac {\sqrt {a}\,\sqrt {d}\,\left (256\,a\,e+256\,b\,d\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}}{\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 {60\,d^2\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {-b\,e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {-b\,e}}-\frac {2\,\ln \left (\frac {\sqrt {e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {d+e\,x}-\sqrt {d}}-\sqrt {b}\right )\,\left (3\,a^2\,e^2+2\,a\,b\,d\,e+3\,b^2\,d^2\right )}{b^{5/2}\,\sqrt {e}}+\frac {\ln \left (\sqrt {b}+\frac {\sqrt {e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {d+e\,x}-\sqrt {d}}\right )\,\left (6\,a^2\,e^2+4\,a\,b\,d\,e+6\,b^2\,d^2\right )}{b^{5/2}\,\sqrt {e}}-\frac {40\,d\,\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*d^2 + 8*e^2*x^2 + 20*d*e*x)/((a + b*x)^(1/2)*(d + e*x)^(1/2)),x)

[Out]

(((40*b*d^2 + 40*a*d*e)*((a + b*x)^(1/2) - a^(1/2)))/(e^2*((d + e*x)^(1/2) - d^(1/2))) - (160*a^(1/2)*d^(3/2)*
((a + b*x)^(1/2) - a^(1/2))^2)/(e*((d + e*x)^(1/2) - d^(1/2))^2) + ((40*b*d^2 + 40*a*d*e)*((a + b*x)^(1/2) - a
^(1/2))^3)/(b*e*((d + e*x)^(1/2) - d^(1/2))^3))/(((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)) - ((((a + b*x)^(1/2) - a^(1/
2))*(12*b^3*d^2 + 12*a^2*b*e^2 + 8*a*b^2*d*e))/(e^4*((d + e*x)^(1/2) - d^(1/2))) - (((a + b*x)^(1/2) - a^(1/2)
)^3*(44*a^2*e^2 + 44*b^2*d^2 + 200*a*b*d*e))/(e^3*((d + e*x)^(1/2) - d^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2)
)^7*(12*a^2*e^2 + 12*b^2*d^2 + 8*a*b*d*e))/(b^2*e*((d + e*x)^(1/2) - d^(1/2))^7) - (((a + b*x)^(1/2) - a^(1/2)
)^5*(44*a^2*e^2 + 44*b^2*d^2 + 200*a*b*d*e))/(b*e^2*((d + e*x)^(1/2) - d^(1/2))^5) + (a^(1/2)*d^(1/2)*(256*a*e
+ 256*b*d)*((a + b*x)^(1/2) - a^(1/2))^4)/(e^2*((d + e*x)^(1/2) - d^(1/2))^4))/(((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)) - (60*d^2*atan((b*((d + e*x)^(1/2) - d^(1/2)))/((-b*e)^(1/2)*((a
+ b*x)^(1/2) - a^(1/2)))))/(-b*e)^(1/2) - (2*log((e^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/((d + e*x)^(1/2) - d^(
1/2)) - b^(1/2))*(3*a^2*e^2 + 3*b^2*d^2 + 2*a*b*d*e))/(b^(5/2)*e^(1/2)) + (log(b^(1/2) + (e^(1/2)*((a + b*x)^(
1/2) - a^(1/2)))/((d + e*x)^(1/2) - d^(1/2)))*(6*a^2*e^2 + 6*b^2*d^2 + 4*a*b*d*e))/(b^(5/2)*e^(1/2)) - (40*d*a
tanh((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
))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {15 d^{2} + 20 d e x + 8 e^{2} x^{2}}{\sqrt {a + b x} \sqrt {d + e x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e**2*x**2+20*d*e*x+15*d**2)/(e*x+d)**(1/2)/(b*x+a)**(1/2),x)

[Out]

Integral((15*d**2 + 20*d*e*x + 8*e**2*x**2)/(sqrt(a + b*x)*sqrt(d + e*x)), x)

________________________________________________________________________________________