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

Optimal. Leaf size=196 \[ -\frac {(b d-a e)^2 (2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{5/2} e^{5/2}}+\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (2 A b e-B (a e+b d))}{8 b^2 e^2}+\frac {(a+b x)^{3/2} \sqrt {d+e x} (2 A b e-B (a e+b d))}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e} \]

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Rubi [A]  time = 0.15, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (2 A b e-B (a e+b d))}{8 b^2 e^2}-\frac {(b d-a e)^2 (2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{5/2} e^{5/2}}+\frac {(a+b x)^{3/2} \sqrt {d+e x} (2 A b e-B (a e+b d))}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)*(2*A*b*e - B*(b*d + a*e))*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^2*e^2) + ((2*A*b*e - B*(b*d + a*e))*(
a + b*x)^(3/2)*Sqrt[d + e*x])/(4*b^2*e) + (B*(a + b*x)^(3/2)*(d + e*x)^(3/2))/(3*b*e) - ((b*d - a*e)^2*(2*A*b*
e - B*(b*d + a*e))*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(5/2)*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 \sqrt {a+b x} (A+B x) \sqrt {d+e x} \, dx &=\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}+\frac {\left (3 A b e-B \left (\frac {3 b d}{2}+\frac {3 a e}{2}\right )\right ) \int \sqrt {a+b x} \sqrt {d+e x} \, dx}{3 b e}\\ &=\frac {(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt {d+e x}}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}+\frac {\left ((b d-a e) \left (3 A b e-B \left (\frac {3 b d}{2}+\frac {3 a e}{2}\right )\right )\right ) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}} \, dx}{12 b^2 e}\\ &=\frac {(b d-a e) (2 A b e-B (b d+a e)) \sqrt {a+b x} \sqrt {d+e x}}{8 b^2 e^2}+\frac {(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt {d+e x}}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac {\left ((b d-a e)^2 \left (3 A b e-B \left (\frac {3 b d}{2}+\frac {3 a e}{2}\right )\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{24 b^2 e^2}\\ &=\frac {(b d-a e) (2 A b e-B (b d+a e)) \sqrt {a+b x} \sqrt {d+e x}}{8 b^2 e^2}+\frac {(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt {d+e x}}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac {\left ((b d-a e)^2 \left (3 A b e-B \left (\frac {3 b d}{2}+\frac {3 a e}{2}\right )\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 )}{12 b^3 e^2}\\ &=\frac {(b d-a e) (2 A b e-B (b d+a e)) \sqrt {a+b x} \sqrt {d+e x}}{8 b^2 e^2}+\frac {(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt {d+e x}}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac {\left ((b d-a e)^2 \left (3 A b e-B \left (\frac {3 b d}{2}+\frac {3 a e}{2}\right )\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 )}{12 b^3 e^2}\\ &=\frac {(b d-a e) (2 A b e-B (b d+a e)) \sqrt {a+b x} \sqrt {d+e x}}{8 b^2 e^2}+\frac {(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt {d+e x}}{4 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac {(b d-a e)^2 (2 A b e-B (b d+a e)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{5/2} e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 1.30, size = 201, normalized size = 1.03 \begin {gather*} \frac {(a+b x)^{3/2} (d+e x)^{3/2} \left (3 B-\frac {9 (a B e-2 A b e+b B d) \left (\sqrt {e} (a+b x) \sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}} (a e+b (d+2 e x))-\sqrt {a+b x} (b d-a e)^2 \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )\right )}{8 e^{3/2} (a+b x)^2 (b d-a e)^{3/2} \left (\frac {b (d+e x)}{b d-a e}\right )^{3/2}}\right )}{9 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(3/2)*(d + e*x)^(3/2)*(3*B - (9*(b*B*d - 2*A*b*e + a*B*e)*(Sqrt[e]*Sqrt[b*d - a*e]*(a + b*x)*Sqrt[(
b*(d + e*x))/(b*d - a*e)]*(a*e + b*(d + 2*e*x)) - (b*d - a*e)^2*Sqrt[a + b*x]*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/
Sqrt[b*d - a*e]]))/(8*e^(3/2)*(b*d - a*e)^(3/2)*(a + b*x)^2*((b*(d + e*x))/(b*d - a*e))^(3/2))))/(9*b*e)

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IntegrateAlgebraic [A]  time = 0.45, size = 254, normalized size = 1.30 \begin {gather*} \frac {(b d-a e)^2 (a B e-2 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{8 b^{5/2} e^{5/2}}+\frac {\sqrt {d+e x} (b d-a e)^2 \left (\frac {6 A b^3 e (d+e x)^2}{(a+b x)^2}-\frac {3 b^3 B d (d+e x)^2}{(a+b x)^2}-\frac {3 a b^2 B e (d+e x)^2}{(a+b x)^2}+\frac {8 b^2 B d e (d+e x)}{a+b x}-\frac {8 a b B e^2 (d+e x)}{a+b x}+3 a B e^3-6 A b e^3+3 b B d e^2\right )}{24 b^2 e^2 \sqrt {a+b x} \left (\frac {b (d+e x)}{a+b x}-e\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^2*Sqrt[d + e*x]*(3*b*B*d*e^2 - 6*A*b*e^3 + 3*a*B*e^3 + (8*b^2*B*d*e*(d + e*x))/(a + b*x) - (8*a*b
*B*e^2*(d + e*x))/(a + b*x) - (3*b^3*B*d*(d + e*x)^2)/(a + b*x)^2 + (6*A*b^3*e*(d + e*x)^2)/(a + b*x)^2 - (3*a
*b^2*B*e*(d + e*x)^2)/(a + b*x)^2))/(24*b^2*e^2*Sqrt[a + b*x]*(-e + (b*(d + e*x))/(a + b*x))^3) + ((b*d - a*e)
^2*(b*B*d - 2*A*b*e + a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/(8*b^(5/2)*e^(5/2))

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fricas [A]  time = 1.41, size = 528, normalized size = 2.69 \begin {gather*} \left [-\frac {3 \, {\left (B b^{3} d^{3} - {\left (B a b^{2} + 2 \, A b^{3}\right )} d^{2} e - {\left (B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 2 \, A a^{2} b\right )} 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 b^{3} e^{3} x^{2} - 3 \, B b^{3} d^{2} e + 2 \, {\left (B a b^{2} + 3 \, A b^{3}\right )} d e^{2} - 3 \, {\left (B a^{2} b - 2 \, A a b^{2}\right )} e^{3} + 2 \, {\left (B b^{3} d e^{2} + {\left (B a b^{2} + 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{96 \, b^{3} e^{3}}, -\frac {3 \, {\left (B b^{3} d^{3} - {\left (B a b^{2} + 2 \, A b^{3}\right )} d^{2} e - {\left (B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 2 \, A a^{2} b\right )} 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 b^{3} e^{3} x^{2} - 3 \, B b^{3} d^{2} e + 2 \, {\left (B a b^{2} + 3 \, A b^{3}\right )} d e^{2} - 3 \, {\left (B a^{2} b - 2 \, A a b^{2}\right )} e^{3} + 2 \, {\left (B b^{3} d e^{2} + {\left (B a b^{2} + 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{48 \, b^{3} 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]

[-1/96*(3*(B*b^3*d^3 - (B*a*b^2 + 2*A*b^3)*d^2*e - (B*a^2*b - 4*A*a*b^2)*d*e^2 + (B*a^3 - 2*A*a^2*b)*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*b^3*e^3*x^2 - 3*B*b^3*d^2*e + 2*(B*a*b^2 + 3*A*b^3)*d*e^2 - 3*(B
*a^2*b - 2*A*a*b^2)*e^3 + 2*(B*b^3*d*e^2 + (B*a*b^2 + 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^3),
-1/48*(3*(B*b^3*d^3 - (B*a*b^2 + 2*A*b^3)*d^2*e - (B*a^2*b - 4*A*a*b^2)*d*e^2 + (B*a^3 - 2*A*a^2*b)*e^3)*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*(8*B*b^3*e^3*x^2 - 3*B*b^3*d^2*e + 2*(B*a*b^2 + 3*A*b^3)*d*e^2 - 3*(B*a^2*b - 2*A*a*b^2)
*e^3 + 2*(B*b^3*d*e^2 + (B*a*b^2 + 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^3)]

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giac [B]  time = 1.76, size = 571, normalized size = 2.91 \begin {gather*} -\frac {\frac {24 \, {\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 a {\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 )} B {\left | b \right |}}{b} - \frac {6 \, {\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 a {\left | b \right |}}{b^{3}} - \frac {6 \, {\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 )} A {\left | b \right |}}{b^{2}}}{24 \, b} \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]

-1/24*(24*((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*a*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))*B*abs(b)/b - 6*((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*a*abs(b)/b^3 - 6*((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))*A*abs(b)/b^2)/b

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maple [B]  time = 0.02, size = 755, normalized size = 3.85 \begin {gather*} -\frac {\sqrt {b x +a}\, \sqrt {e x +d}\, \left (6 A \,a^{2} b \,e^{3} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-12 A a \,b^{2} d \,e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+6 A \,b^{3} d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-3 B \,a^{3} e^{3} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+3 B \,a^{2} b d \,e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+3 B a \,b^{2} d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-3 B \,b^{3} d^{3} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-16 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B \,b^{2} e^{2} x^{2}-24 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, A \,b^{2} e^{2} x -4 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B a b \,e^{2} x -4 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B \,b^{2} d e x -12 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, A a b \,e^{2}-12 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, A \,b^{2} d e +6 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B \,a^{2} e^{2}-4 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B a b d e +6 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B \,b^{2} d^{2}\right )}{48 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, b^{2} 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]

-1/48*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(-16*B*x^2*b^2*e^2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+6*A*ln(1/2*(2
*b*e*x+a*e+b*d+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a^2*b*e^3-12*A*ln(1/2*(2*b*e*x+a*e+
b*d+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a*b^2*d*e^2+6*A*ln(1/2*(2*b*e*x+a*e+b*d+2*(b*e
*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*b^3*d^2*e-24*A*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/
2)*x*b^2*e^2-3*B*ln(1/2*(2*b*e*x+a*e+b*d+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a^3*e^3+3
*B*ln(1/2*(2*b*e*x+a*e+b*d+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a^2*b*d*e^2+3*B*ln(1/2*
(2*b*e*x+a*e+b*d+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a*b^2*d^2*e-3*B*ln(1/2*(2*b*e*x+a
*e+b*d+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*b^3*d^3-4*B*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)
*(b*e)^(1/2)*x*a*b*e^2-4*B*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*x*b^2*d*e-12*A*(b*e*x^2+a*e*x+b*d*x+a*d
)^(1/2)*(b*e)^(1/2)*a*b*e^2-12*A*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*b^2*d*e+6*B*(b*e*x^2+a*e*x+b*d*x+
a*d)^(1/2)*(b*e)^(1/2)*a^2*e^2-4*B*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*a*b*d*e+6*B*(b*e*x^2+a*e*x+b*d*
x+a*d)^(1/2)*(b*e)^(1/2)*b^2*d^2)/(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)/b^2/e^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((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 = 79.38, size = 1207, normalized size = 6.16

result too large to display

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*(x/2 + (a*e + b*d)/(4*b*e))*(a + b*x)^(1/2)*(d + e*x)^(1/2) - ((((a + b*x)^(1/2) - a^(1/2))*((B*b^6*d^3)/4 +
(B*a^3*b^3*e^3)/4 - (B*a^2*b^4*d*e^2)/4 - (B*a*b^5*d^2*e)/4))/(e^8*((d + e*x)^(1/2) - d^(1/2))) - (((a + b*x)
^(1/2) - a^(1/2))^3*((17*B*b^5*d^3)/12 + (17*B*a^3*b^2*e^3)/12 + (101*B*a^2*b^3*d*e^2)/4 + (101*B*a*b^4*d^2*e)
/4))/(e^7*((d + e*x)^(1/2) - d^(1/2))^3) - (((a + b*x)^(1/2) - a^(1/2))^7*((19*B*a^3*e^3)/2 + (19*B*b^3*d^3)/2
+ (269*B*a*b^2*d^2*e)/2 + (269*B*a^2*b*d*e^2)/2))/(e^5*((d + e*x)^(1/2) - d^(1/2))^7) - (((a + b*x)^(1/2) - a
^(1/2))^5*((19*B*b^4*d^3)/2 + (19*B*a^3*b*e^3)/2 + (269*B*a^2*b^2*d*e^2)/2 + (269*B*a*b^3*d^2*e)/2))/(e^6*((d
+ e*x)^(1/2) - d^(1/2))^5) + (((a + b*x)^(1/2) - a^(1/2))^11*((B*a^3*e^3)/4 + (B*b^3*d^3)/4 - (B*a*b^2*d^2*e)/
4 - (B*a^2*b*d*e^2)/4))/(b^2*e^3*((d + e*x)^(1/2) - d^(1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^9*((17*B*a^3*e
^3)/12 + (17*B*b^3*d^3)/12 + (101*B*a*b^2*d^2*e)/4 + (101*B*a^2*b*d*e^2)/4))/(b*e^4*((d + e*x)^(1/2) - d^(1/2)
)^9) + (a^(1/2)*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^4*(32*B*b^4*d^2 + 32*B*a^2*b^2*e^2 + 96*B*a*b^3*d*e))/(e^6
*((d + e*x)^(1/2) - d^(1/2))^4) + (8*B*a^(3/2)*d^(3/2)*((a + b*x)^(1/2) - a^(1/2))^10)/(e^2*((d + e*x)^(1/2) -
d^(1/2))^10) + (a^(1/2)*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(64*B*b^3*d^2 + 64*B*a^2*b*e^2 + (656*B*a*b^2*d
*e)/3))/(e^5*((d + e*x)^(1/2) - d^(1/2))^6) + (a^(1/2)*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^8*(32*B*a^2*e^2 + 3
2*B*b^2*d^2 + 96*B*a*b*d*e))/(e^4*((d + e*x)^(1/2) - d^(1/2))^8) + (8*B*a^(3/2)*b^4*d^(3/2)*((a + b*x)^(1/2) -
a^(1/2))^2)/(e^6*((d + e*x)^(1/2) - d^(1/2))^2))/(((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)) - (A*log(a*e + b*d + 2*b*e*x + 2*b^(1/2)*e^(1
/2)*(a + b*x)^(1/2)*(d + e*x)^(1/2))*(a*e - b*d)^2)/(8*b^(3/2)*e^(3/2)) + (B*atanh((B*e^(1/2)*(a*e + b*d)*(a*e
- b*d)^2*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((d + e*x)^(1/2) - d^(1/2))*(B*a^3*e^3 + B*b^3*d^3 - B*a*b^2*d
^2*e - B*a^2*b*d*e^2)))*(a*e + b*d)*(a*e - b*d)^2)/(4*b^(5/2)*e^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + B x\right ) \sqrt {a + b x} \sqrt {d + e x}\, dx \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]

Integral((A + B*x)*sqrt(a + b*x)*sqrt(d + e*x), x)

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