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

Optimal. Leaf size=176 \[ \frac {(b d-a e) \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{7/2} \sqrt {e}}+\frac {\sqrt {a+b x} \sqrt {d+e x} \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right )}{b^3}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {2 \sqrt {a+b x} (d+e x)^{3/2} (4 b d-3 a e)}{b^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {951, 80, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x} \sqrt {d+e x} \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right )}{b^3}+\frac {(b d-a e) \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{7/2} \sqrt {e}}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {2 \sqrt {a+b x} (d+e x)^{3/2} (4 b d-3 a e)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((11*b^2*d^2 - 13*a*b*d*e + 5*a^2*e^2)*Sqrt[a + b*x]*Sqrt[d + e*x])/b^3 + (2*(4*b*d - 3*a*e)*Sqrt[a + b*x]*(d
+ e*x)^(3/2))/b^2 + (8*e*(a + b*x)^(3/2)*(d + e*x)^(3/2))/(3*b^2) + ((b*d - a*e)*(11*b^2*d^2 - 13*a*b*d*e + 5*
a^2*e^2)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(7/2)*Sqrt[e])

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]

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 {\sqrt {d+e x} \left (15 d^2+20 d e x+8 e^2 x^2\right )}{\sqrt {a+b x}} \, dx &=\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {\int \frac {\sqrt {d+e x} \left (3 e (3 b d-2 a e) (5 b d+2 a e)+12 b e^2 (4 b d-3 a e) x\right )}{\sqrt {a+b x}} \, dx}{3 b^2 e}\\ &=\frac {2 (4 b d-3 a e) \sqrt {a+b x} (d+e x)^{3/2}}{b^2}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x}} \, dx}{b^2}\\ &=\frac {\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt {a+b x} \sqrt {d+e x}}{b^3}+\frac {2 (4 b d-3 a e) \sqrt {a+b x} (d+e x)^{3/2}}{b^2}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {\left ((b d-a e) \left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{2 b^3}\\ &=\frac {\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt {a+b x} \sqrt {d+e x}}{b^3}+\frac {2 (4 b d-3 a e) \sqrt {a+b x} (d+e x)^{3/2}}{b^2}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {\left ((b d-a e) \left (11 b^2 d^2-13 a b d e+5 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^4}\\ &=\frac {\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt {a+b x} \sqrt {d+e x}}{b^3}+\frac {2 (4 b d-3 a e) \sqrt {a+b x} (d+e x)^{3/2}}{b^2}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {\left ((b d-a e) \left (11 b^2 d^2-13 a b d e+5 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^4}\\ &=\frac {\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt {a+b x} \sqrt {d+e x}}{b^3}+\frac {2 (4 b d-3 a e) \sqrt {a+b x} (d+e x)^{3/2}}{b^2}+\frac {8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac {(b d-a e) \left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{7/2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 163, normalized size = 0.93 \begin {gather*} \frac {\sqrt {d+e x} \left (\sqrt {a+b x} \left (15 a^2 e^2-a b e (49 d+10 e x)+b^2 \left (57 d^2+32 d e x+8 e^2 x^2\right )\right )+\frac {3 \sqrt {b d-a e} \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right ) \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{\sqrt {e} \sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(Sqrt[a + b*x]*(15*a^2*e^2 - a*b*e*(49*d + 10*e*x) + b^2*(57*d^2 + 32*d*e*x + 8*e^2*x^2)) + (3*
Sqrt[b*d - a*e]*(11*b^2*d^2 - 13*a*b*d*e + 5*a^2*e^2)*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/(Sqrt[
e]*Sqrt[(b*(d + e*x))/(b*d - a*e)])))/(3*b^3)

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IntegrateAlgebraic [A]  time = 0.67, size = 220, normalized size = 1.25 \begin {gather*} \frac {\sqrt {a+\frac {b (d+e x)}{e}-\frac {b d}{e}} \left (15 a^2 e^2 \sqrt {d+e x}-10 a b e (d+e x)^{3/2}-39 a b d e \sqrt {d+e x}+33 b^2 d^2 \sqrt {d+e x}+8 b^2 (d+e x)^{5/2}+16 b^2 d (d+e x)^{3/2}\right )}{3 b^3}-\frac {\sqrt {\frac {b}{e}} \left (-5 a^3 e^3+18 a^2 b d e^2-24 a b^2 d^2 e+11 b^3 d^3\right ) \log \left (\sqrt {a+\frac {b (d+e x)}{e}-\frac {b d}{e}}-\sqrt {\frac {b}{e}} \sqrt {d+e x}\right )}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - (b*d)/e + (b*(d + e*x))/e]*(33*b^2*d^2*Sqrt[d + e*x] - 39*a*b*d*e*Sqrt[d + e*x] + 15*a^2*e^2*Sqrt[d
+ e*x] + 16*b^2*d*(d + e*x)^(3/2) - 10*a*b*e*(d + e*x)^(3/2) + 8*b^2*(d + e*x)^(5/2)))/(3*b^3) - (Sqrt[b/e]*(1
1*b^3*d^3 - 24*a*b^2*d^2*e + 18*a^2*b*d*e^2 - 5*a^3*e^3)*Log[-(Sqrt[b/e]*Sqrt[d + e*x]) + Sqrt[a - (b*d)/e + (
b*(d + e*x))/e]])/b^4

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fricas [A]  time = 0.46, size = 414, normalized size = 2.35 \begin {gather*} \left [-\frac {3 \, {\left (11 \, b^{3} d^{3} - 24 \, a b^{2} d^{2} e + 18 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\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 (8 \, b^{3} e^{3} x^{2} + 57 \, b^{3} d^{2} e - 49 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{12 \, b^{4} e}, -\frac {3 \, {\left (11 \, b^{3} d^{3} - 24 \, a b^{2} d^{2} e + 18 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\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 (8 \, b^{3} e^{3} x^{2} + 57 \, b^{3} d^{2} e - 49 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, b^{4} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(11*b^3*d^3 - 24*a*b^2*d^2*e + 18*a^2*b*d*e^2 - 5*a^3*e^3)*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*(8*b^3*e^3*x^2 + 57*b^3*d^2*e - 49*a*b^2*d*e^2 + 15*a^2*b*e^3 + 2*(16*b^3*d*e^2 - 5*a*b^2*e^3)*x)*sqrt(b*x
+ a)*sqrt(e*x + d))/(b^4*e), -1/6*(3*(11*b^3*d^3 - 24*a*b^2*d^2*e + 18*a^2*b*d*e^2 - 5*a^3*e^3)*sqrt(-b*e)*ar
ctan(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*(8*b^3*e^3*x^2 + 57*b^3*d^2*e - 49*a*b^2*d*e^2 + 15*a^2*b*e^3 + 2*(16*b^3*d*e^2 - 5*a*b^2*e^3)*x)
*sqrt(b*x + a)*sqrt(e*x + d))/(b^4*e)]

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giac [B]  time = 0.38, size = 441, normalized size = 2.51 \begin {gather*} -\frac {\frac {45 \, {\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 )} d^{2} {\left | b \right |}}{b^{2}} - \frac {{\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {{\left (b^{6} d e^{3} - 13 \, a b^{5} e^{4}\right )} e^{\left (-4\right )}}{b^{7}}\right )} - \frac {3 \, {\left (b^{7} d^{2} e^{2} + 2 \, a b^{6} d e^{3} - 11 \, a^{2} b^{5} e^{4}\right )} e^{\left (-4\right )}}{b^{7}}\right )} - \frac {3 \, {\left (b^{3} d^{3} + a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\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 )} {\left | b \right |} e^{2}}{b^{2}} - \frac {15 \, {\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 )} d {\left | b \right |} e}{b^{3}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(45*((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))*d^2*abs(b)/b^2 - (sqrt(b^2*d + (b*x + a)*b*e
- a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)*e^(-4)/b^7) - 3*(b^7*d^2*e^2
+ 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)*e^(-4)/b^7) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*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))*abs(b)*e^2/b^2 - 15*
((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))*d*abs(b)*e/b^3)/b

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maple [B]  time = 0.02, size = 392, normalized size = 2.23 \begin {gather*} -\frac {\sqrt {e x +d}\, \sqrt {b x +a}\, \left (15 a^{3} e^{3} \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 )-54 a^{2} b d \,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 )+72 a \,b^{2} d^{2} 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 )-33 b^{3} d^{3} \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 )-16 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b^{2} e^{2} x^{2}+20 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a b \,e^{2} x -64 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} d e x -30 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a^{2} e^{2}+98 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a b d e -114 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} d^{2}\right )}{6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(-16*x^2*b^2*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+15*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^3*e^3-54*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*b*d*e^2+72*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^2*d^2*e-33*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^3*d^
3+20*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*x*a*b*e^2-64*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*x*b^2*d*e-30*(b*e)^(
1/2)*((b*x+a)*(e*x+d))^(1/2)*a^2*e^2+98*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*d*e-114*(b*e)^(1/2)*((b*x+a)*(
e*x+d))^(1/2)*b^2*d^2)/b^3/((b*x+a)*(e*x+d))^(1/2)/(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((e*x+d)^(1/2)*(8*e^2*x^2+20*d*e*x+15*d^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 = 73.15, size = 1797, normalized size = 10.21

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x)^(1/2) - a^(1/2))^3*(70*b^2*d^3 + 110*a^2*d*e^2 + 460*a*b*d^2*e))/(e^3*((d + e*x)^(1/2) - d^(1/2))
^3) + (((a + b*x)^(1/2) - a^(1/2))*(10*b^3*d^3 + 20*a*b^2*d^2*e - 30*a^2*b*d*e^2))/(e^4*((d + e*x)^(1/2) - d^(
1/2))) - (160*a^(1/2)*d^(5/2)*((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))^7*(10*b^2*d^3 - 30*a^2*d*e^2 + 20*a*b*d^2*e))/(b^2*e*((d + e*x)^(1/2) - d^(1/2))^7) + (((a + b
*x)^(1/2) - a^(1/2))^5*(70*b^2*d^3 + 110*a^2*d*e^2 + 460*a*b*d^2*e))/(b*e^2*((d + e*x)^(1/2) - d^(1/2))^5) - (
a^(1/2)*d^(1/2)*(320*b*d^2 + 640*a*d*e)*((a + b*x)^(1/2) - a^(1/2))^4)/(e^2*((d + e*x)^(1/2) - d^(1/2))^4) - (
160*a^(1/2)*b^2*d^(5/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(e^3*((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*b^5*d^3 - 10*a^3*
b^2*e^3 + 6*a^2*b^3*d*e^2 + 2*a*b^4*d^2*e))/(e^6*((d + e*x)^(1/2) - d^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^5
*(132*a^3*e^3 + 76*b^3*d^3 + 1100*a*b^2*d^2*e + 1252*a^2*b*d*e^2))/(e^4*((d + e*x)^(1/2) - d^(1/2))^5) - (((a
+ b*x)^(1/2) - a^(1/2))^3*((34*b^4*d^3)/3 - (170*a^3*b*e^3)/3 + 34*a^2*b^2*d*e^2 + 182*a*b^3*d^2*e))/(e^5*((d
+ e*x)^(1/2) - d^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))^11*(2*b^3*d^3 - 10*a^3*e^3 + 2*a*b^2*d^2*e + 6*a^2*b
*d*e^2))/(b^3*e*((d + e*x)^(1/2) - d^(1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^9*((34*b^3*d^3)/3 - (170*a^3*e^
3)/3 + 182*a*b^2*d^2*e + 34*a^2*b*d*e^2))/(b^2*e^2*((d + e*x)^(1/2) - d^(1/2))^9) - (((a + b*x)^(1/2) - a^(1/2
))^7*(132*a^3*e^3 + 76*b^3*d^3 + 1100*a*b^2*d^2*e + 1252*a^2*b*d*e^2))/(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))^6*(1024*a^2*e^2 + 512*b^2*d^2 + (5632*a*b*d*e)/3))/(e^3*((d + e*
x)^(1/2) - d^(1/2))^6) + (a^(1/2)*d^(1/2)*(256*b*d^2 + 768*a*d*e)*((a + b*x)^(1/2) - a^(1/2))^8)/(e^2*((d + e*
x)^(1/2) - d^(1/2))^8) + (a^(1/2)*d^(1/2)*(256*b^3*d^2 + 768*a*b^2*d*e)*((a + b*x)^(1/2) - a^(1/2))^4)/(e^4*((
d + e*x)^(1/2) - d^(1/2))^4))/(((a + b*x)^(1/2) - a^(1/2))^12/((d + e*x)^(1/2) - d^(1/2))^12 + b^6/e^6 - (6*b^
5*((a + b*x)^(1/2) - a^(1/2))^2)/(e^5*((d + e*x)^(1/2) - d^(1/2))^2) + (15*b^4*((a + b*x)^(1/2) - a^(1/2))^4)/
(e^4*((d + e*x)^(1/2) - d^(1/2))^4) - (20*b^3*((a + b*x)^(1/2) - a^(1/2))^6)/(e^3*((d + e*x)^(1/2) - d^(1/2))^
6) + (15*b^2*((a + b*x)^(1/2) - a^(1/2))^8)/(e^2*((d + e*x)^(1/2) - d^(1/2))^8) - (6*b*((a + b*x)^(1/2) - a^(1
/2))^10)/(e*((d + e*x)^(1/2) - d^(1/2))^10)) + (((30*b*d^3 + 30*a*d^2*e)*((a + b*x)^(1/2) - a^(1/2)))/(e^2*((d
+ e*x)^(1/2) - d^(1/2))) - (120*a^(1/2)*d^(5/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(e*((d + e*x)^(1/2) - d^(1/2))
^2) + ((30*b*d^3 + 30*a*d^2*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)) - (2*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)*(5*a^2*e^2 + b^2*d^2 + 2*a*b*d*e))/(b^(7/2)*e^(1/2)) - (30*d^2*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)) + (10*d*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))/(b^(5/2)*e^(1/2))

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

[Out]

Timed out

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