3.2213 \(\int (a+b x)^{3/2} (A+B x) \sqrt {d+e x} \, dx\)

Optimal. Leaf size=250 \[ -\frac {(b d-a e)^3 (3 a B e-8 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{64 b^{5/2} e^{7/2}}+\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e)^2 (3 a B e-8 A b e+5 b B d)}{64 b^2 e^3}-\frac {(a+b x)^{3/2} \sqrt {d+e x} (b d-a e) (3 a B e-8 A b e+5 b B d)}{96 b^2 e^2}-\frac {(a+b x)^{5/2} \sqrt {d+e x} (3 a B e-8 A b e+5 b B d)}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e} \]

[Out]

1/4*B*(b*x+a)^(5/2)*(e*x+d)^(3/2)/b/e-1/64*(-a*e+b*d)^3*(-8*A*b*e+3*B*a*e+5*B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1/
2)/b^(1/2)/(e*x+d)^(1/2))/b^(5/2)/e^(7/2)-1/96*(-a*e+b*d)*(-8*A*b*e+3*B*a*e+5*B*b*d)*(b*x+a)^(3/2)*(e*x+d)^(1/
2)/b^2/e^2-1/24*(-8*A*b*e+3*B*a*e+5*B*b*d)*(b*x+a)^(5/2)*(e*x+d)^(1/2)/b^2/e+1/64*(-a*e+b*d)^2*(-8*A*b*e+3*B*a
*e+5*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/b^2/e^3

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Rubi [A]  time = 0.19, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e)^2 (3 a B e-8 A b e+5 b B d)}{64 b^2 e^3}-\frac {(a+b x)^{3/2} \sqrt {d+e x} (b d-a e) (3 a B e-8 A b e+5 b B d)}{96 b^2 e^2}-\frac {(b d-a e)^3 (3 a B e-8 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{64 b^{5/2} e^{7/2}}-\frac {(a+b x)^{5/2} \sqrt {d+e x} (3 a B e-8 A b e+5 b B d)}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)*(A + B*x)*Sqrt[d + e*x],x]

[Out]

((b*d - a*e)^2*(5*b*B*d - 8*A*b*e + 3*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(64*b^2*e^3) - ((b*d - a*e)*(5*b*B*d
- 8*A*b*e + 3*a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(96*b^2*e^2) - ((5*b*B*d - 8*A*b*e + 3*a*B*e)*(a + b*x)^(
5/2)*Sqrt[d + e*x])/(24*b^2*e) + (B*(a + b*x)^(5/2)*(d + e*x)^(3/2))/(4*b*e) - ((b*d - a*e)^3*(5*b*B*d - 8*A*b
*e + 3*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(64*b^(5/2)*e^(7/2))

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 (a+b x)^{3/2} (A+B x) \sqrt {d+e x} \, dx &=\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}+\frac {\left (4 A b e-B \left (\frac {5 b d}{2}+\frac {3 a e}{2}\right )\right ) \int (a+b x)^{3/2} \sqrt {d+e x} \, dx}{4 b e}\\ &=-\frac {(5 b B d-8 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}-\frac {((b d-a e) (5 b B d-8 A b e+3 a B e)) \int \frac {(a+b x)^{3/2}}{\sqrt {d+e x}} \, dx}{48 b^2 e}\\ &=-\frac {(b d-a e) (5 b B d-8 A b e+3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{96 b^2 e^2}-\frac {(5 b B d-8 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}+\frac {\left ((b d-a e)^2 (5 b B d-8 A b e+3 a B e)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}} \, dx}{64 b^2 e^2}\\ &=\frac {(b d-a e)^2 (5 b B d-8 A b e+3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{64 b^2 e^3}-\frac {(b d-a e) (5 b B d-8 A b e+3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{96 b^2 e^2}-\frac {(5 b B d-8 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}-\frac {\left ((b d-a e)^3 (5 b B d-8 A b e+3 a B e)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{128 b^2 e^3}\\ &=\frac {(b d-a e)^2 (5 b B d-8 A b e+3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{64 b^2 e^3}-\frac {(b d-a e) (5 b B d-8 A b e+3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{96 b^2 e^2}-\frac {(5 b B d-8 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}-\frac {\left ((b d-a e)^3 (5 b B d-8 A b e+3 a B e)\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 )}{64 b^3 e^3}\\ &=\frac {(b d-a e)^2 (5 b B d-8 A b e+3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{64 b^2 e^3}-\frac {(b d-a e) (5 b B d-8 A b e+3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{96 b^2 e^2}-\frac {(5 b B d-8 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}-\frac {\left ((b d-a e)^3 (5 b B d-8 A b e+3 a B e)\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 )}{64 b^3 e^3}\\ &=\frac {(b d-a e)^2 (5 b B d-8 A b e+3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{64 b^2 e^3}-\frac {(b d-a e) (5 b B d-8 A b e+3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{96 b^2 e^2}-\frac {(5 b B d-8 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{24 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{3/2}}{4 b e}-\frac {(b d-a e)^3 (5 b B d-8 A b e+3 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{64 b^{5/2} e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 2.24, size = 308, normalized size = 1.23 \[ \frac {(a+b x)^{5/2} (d+e x)^{3/2} \left (\frac {5 (-3 a B e+8 A b e-5 b B d) \left (8 b^3 e^3 (a+b x)^3 \sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}}-b (b d-a e) \left (-2 b^2 e^2 (a+b x)^2 \sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}}+3 b^2 e (a+b x) (b d-a e)^{3/2} \sqrt {\frac {b (d+e x)}{b d-a e}}-3 b^2 \sqrt {e} \sqrt {a+b x} (b d-a e)^2 \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )\right )\right )}{48 b^3 e^3 (a+b x)^3 (b d-a e)^{3/2} \left (\frac {b (d+e x)}{b d-a e}\right )^{3/2}}+5 B\right )}{20 b e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)*(A + B*x)*Sqrt[d + e*x],x]

[Out]

((a + b*x)^(5/2)*(d + e*x)^(3/2)*(5*B + (5*(-5*b*B*d + 8*A*b*e - 3*a*B*e)*(8*b^3*e^3*Sqrt[b*d - a*e]*(a + b*x)
^3*Sqrt[(b*(d + e*x))/(b*d - a*e)] - b*(b*d - a*e)*(3*b^2*e*(b*d - a*e)^(3/2)*(a + b*x)*Sqrt[(b*(d + e*x))/(b*
d - a*e)] - 2*b^2*e^2*Sqrt[b*d - a*e]*(a + b*x)^2*Sqrt[(b*(d + e*x))/(b*d - a*e)] - 3*b^2*Sqrt[e]*(b*d - a*e)^
2*Sqrt[a + b*x]*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])))/(48*b^3*e^3*(b*d - a*e)^(3/2)*(a + b*x)^3*
((b*(d + e*x))/(b*d - a*e))^(3/2))))/(20*b*e)

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fricas [A]  time = 0.79, size = 766, normalized size = 3.06 \[ \left [\frac {3 \, {\left (5 \, B b^{4} d^{4} - 4 \, {\left (3 \, B a b^{3} + 2 \, A b^{4}\right )} d^{3} e + 6 \, {\left (B a^{2} b^{2} + 4 \, A a b^{3}\right )} d^{2} e^{2} + 4 \, {\left (B a^{3} b - 6 \, A a^{2} b^{2}\right )} d e^{3} - {\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} e^{4}\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 (48 \, B b^{4} e^{4} x^{3} + 15 \, B b^{4} d^{3} e - {\left (31 \, B a b^{3} + 24 \, A b^{4}\right )} d^{2} e^{2} + {\left (9 \, B a^{2} b^{2} + 64 \, A a b^{3}\right )} d e^{3} - 3 \, {\left (3 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} e^{4} + 8 \, {\left (B b^{4} d e^{3} + {\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} e^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{4} d^{2} e^{2} - 2 \, {\left (5 \, B a b^{3} + 4 \, A b^{4}\right )} d e^{3} - {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} e^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{768 \, b^{3} e^{4}}, \frac {3 \, {\left (5 \, B b^{4} d^{4} - 4 \, {\left (3 \, B a b^{3} + 2 \, A b^{4}\right )} d^{3} e + 6 \, {\left (B a^{2} b^{2} + 4 \, A a b^{3}\right )} d^{2} e^{2} + 4 \, {\left (B a^{3} b - 6 \, A a^{2} b^{2}\right )} d e^{3} - {\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} e^{4}\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 (48 \, B b^{4} e^{4} x^{3} + 15 \, B b^{4} d^{3} e - {\left (31 \, B a b^{3} + 24 \, A b^{4}\right )} d^{2} e^{2} + {\left (9 \, B a^{2} b^{2} + 64 \, A a b^{3}\right )} d e^{3} - 3 \, {\left (3 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} e^{4} + 8 \, {\left (B b^{4} d e^{3} + {\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} e^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{4} d^{2} e^{2} - 2 \, {\left (5 \, B a b^{3} + 4 \, A b^{4}\right )} d e^{3} - {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} e^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{384 \, b^{3} e^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)*(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(5*B*b^4*d^4 - 4*(3*B*a*b^3 + 2*A*b^4)*d^3*e + 6*(B*a^2*b^2 + 4*A*a*b^3)*d^2*e^2 + 4*(B*a^3*b - 6*A*
a^2*b^2)*d*e^3 - (3*B*a^4 - 8*A*a^3*b)*e^4)*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*(48*B*b^4*e^4*x^3 + 1
5*B*b^4*d^3*e - (31*B*a*b^3 + 24*A*b^4)*d^2*e^2 + (9*B*a^2*b^2 + 64*A*a*b^3)*d*e^3 - 3*(3*B*a^3*b - 8*A*a^2*b^
2)*e^4 + 8*(B*b^4*d*e^3 + (9*B*a*b^3 + 8*A*b^4)*e^4)*x^2 - 2*(5*B*b^4*d^2*e^2 - 2*(5*B*a*b^3 + 4*A*b^4)*d*e^3
- (3*B*a^2*b^2 + 56*A*a*b^3)*e^4)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^4), 1/384*(3*(5*B*b^4*d^4 - 4*(3*B*a*
b^3 + 2*A*b^4)*d^3*e + 6*(B*a^2*b^2 + 4*A*a*b^3)*d^2*e^2 + 4*(B*a^3*b - 6*A*a^2*b^2)*d*e^3 - (3*B*a^4 - 8*A*a^
3*b)*e^4)*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*(48*B*b^4*e^4*x^3 + 15*B*b^4*d^3*e - (31*B*a*b^3 + 24*A*b^4)*d^2*e^2 + (9*
B*a^2*b^2 + 64*A*a*b^3)*d*e^3 - 3*(3*B*a^3*b - 8*A*a^2*b^2)*e^4 + 8*(B*b^4*d*e^3 + (9*B*a*b^3 + 8*A*b^4)*e^4)*
x^2 - 2*(5*B*b^4*d^2*e^2 - 2*(5*B*a*b^3 + 4*A*b^4)*d*e^3 - (3*B*a^2*b^2 + 56*A*a*b^3)*e^4)*x)*sqrt(b*x + a)*sq
rt(e*x + d))/(b^3*e^4)]

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giac [B]  time = 2.34, size = 1040, normalized size = 4.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)*(e*x+d)^(1/2),x, algorithm="giac")

[Out]

1/192*(8*(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))*A*abs(b) + (sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 +
(b^12*d*e^5 - 25*a*b^11*e^6)*e^(-6)/b^14) - (5*b^13*d^2*e^4 + 14*a*b^12*d*e^5 - 163*a^2*b^11*e^6)*e^(-6)/b^14)
+ 3*(5*b^14*d^3*e^3 + 9*a*b^13*d^2*e^4 + 15*a^2*b^12*d*e^5 - 93*a^3*b^11*e^6)*e^(-6)/b^14)*sqrt(b*x + a) + 3*
(5*b^4*d^4 + 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 20*a^3*b*d*e^3 - 35*a^4*e^4)*e^(-7/2)*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) - 192*((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^2*abs(b)/b^2 + 16*(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*a*abs(b)/b + 48*((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^2*abs(b)/b^3 + 96*((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*a*abs(b)/b^2)/b

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maple [B]  time = 0.02, size = 1150, normalized size = 4.60 \[ -\frac {\sqrt {b x +a}\, \sqrt {e x +d}\, \left (24 A \,a^{3} b \,e^{4} \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 )-72 A \,a^{2} b^{2} d \,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 )+72 A a \,b^{3} d^{2} 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 )-24 A \,b^{4} d^{3} 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 )-9 B \,a^{4} e^{4} \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 B \,a^{3} b d \,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 )+18 B \,a^{2} b^{2} d^{2} 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 )-36 B a \,b^{3} d^{3} 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 )+15 B \,b^{4} d^{4} \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 )-96 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B \,b^{3} e^{3} x^{3}-128 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, A \,b^{3} e^{3} x^{2}-144 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B a \,b^{2} e^{3} x^{2}-16 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, B \,b^{3} d \,e^{2} x^{2}-224 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, A a \,b^{2} e^{3} x -32 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, A \,b^{3} d \,e^{2} x -12 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B \,a^{2} b \,e^{3} x -40 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B a \,b^{2} d \,e^{2} x +20 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B \,b^{3} d^{2} e x -48 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, A \,a^{2} b \,e^{3}-128 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, A a \,b^{2} d \,e^{2}+48 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, A \,b^{3} d^{2} e +18 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B \,a^{3} e^{3}-18 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B \,a^{2} b d \,e^{2}+62 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B a \,b^{2} d^{2} e -30 \sqrt {b e}\, \sqrt {b e \,x^{2}+a e x +b d x +a d}\, B \,b^{3} d^{3}\right )}{384 \sqrt {b e \,x^{2}+a e x +b d x +a d}\, \sqrt {b e}\, b^{2} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)*(e*x+d)^(1/2),x)

[Out]

-1/384*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(18*B*a^2*b^2*d^2*e^2*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))-24*A*b^4*d^3*e*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))-9*B*a^4*e^4*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))+15*B*b^4*d^4*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))+18*(b*e
)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*B*a^3*e^3-30*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*B*b^3*d^3+24*
A*a^3*b*e^4*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))-72*A*a^2*b^2*d
*e^3*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))+72*A*a*b^3*d^2*e^2*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))+12*B*a^3*b*d*e^3*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))-96*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b
*e)^(1/2)*B*b^3*e^3*x^3-128*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*A*b^3*e^3*x^2-40*(b*e)^(1/2)*(b*e*x^2+
a*e*x+b*d*x+a*d)^(1/2)*B*a*b^2*d*e^2*x-36*B*a*b^3*d^3*e*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))-48*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*A*a^2*b*e^3+48*(b*e)^(1/2)*(b*e*x
^2+a*e*x+b*d*x+a*d)^(1/2)*A*b^3*d^2*e-18*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*B*a^2*b*d*e^2+62*(b*e)^(1
/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*B*a*b^2*d^2*e-224*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*A*a*b^2*e^3*
x-32*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*A*b^3*d*e^2*x-12*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*
B*a^2*b*e^3*x+20*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*B*b^3*d^2*e*x-144*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)
*(b*e)^(1/2)*B*a*b^2*e^3*x^2-16*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*B*b^3*d*e^2*x^2-128*(b*e*x^2+a*e*x
+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*A*a*b^2*d*e^2)/b^2/(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)/e^3/(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)^(3/2)*(B*x+A)*(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 [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}\,\sqrt {d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)*(a + b*x)^(3/2)*(d + e*x)^(1/2),x)

[Out]

int((A + B*x)*(a + b*x)^(3/2)*(d + e*x)^(1/2), x)

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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)**(3/2)*(B*x+A)*(e*x+d)**(1/2),x)

[Out]

Timed out

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