3.2214 \(\int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=193 \[ -\frac {(b d-a e)^2 (a B e-6 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{3/2} e^{7/2}}+\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (a B e-6 A b e+5 b B d)}{8 b e^3}-\frac {(a+b x)^{3/2} \sqrt {d+e x} (a B e-6 A b e+5 b B d)}{12 b e^2}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e} \]

[Out]

-1/8*(-a*e+b*d)^2*(-6*A*b*e+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^(3/2)/e^(7/2
)-1/12*(-6*A*b*e+B*a*e+5*B*b*d)*(b*x+a)^(3/2)*(e*x+d)^(1/2)/b/e^2+1/3*B*(b*x+a)^(5/2)*(e*x+d)^(1/2)/b/e+1/8*(-
a*e+b*d)*(-6*A*b*e+B*a*e+5*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/b/e^3

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Rubi [A]  time = 0.15, antiderivative size = 193, 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} \[ -\frac {(b d-a e)^2 (a B e-6 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{3/2} e^{7/2}}-\frac {(a+b x)^{3/2} \sqrt {d+e x} (a B e-6 A b e+5 b B d)}{12 b e^2}+\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (a B e-6 A b e+5 b B d)}{8 b e^3}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)*(5*b*B*d - 6*A*b*e + a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b*e^3) - ((5*b*B*d - 6*A*b*e + a*B*e)
*(a + b*x)^(3/2)*Sqrt[d + e*x])/(12*b*e^2) + (B*(a + b*x)^(5/2)*Sqrt[d + e*x])/(3*b*e) - ((b*d - a*e)^2*(5*b*B
*d - 6*A*b*e + a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(3/2)*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 \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {d+e x}} \, dx &=\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e}+\frac {\left (3 A b e-B \left (\frac {5 b d}{2}+\frac {a e}{2}\right )\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {d+e x}} \, dx}{3 b e}\\ &=-\frac {(5 b B d-6 A b e+a B e) (a+b x)^{3/2} \sqrt {d+e x}}{12 b e^2}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e}+\frac {((b d-a e) (5 b B d-6 A b e+a B e)) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}} \, dx}{8 b e^2}\\ &=\frac {(b d-a e) (5 b B d-6 A b e+a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b e^3}-\frac {(5 b B d-6 A b e+a B e) (a+b x)^{3/2} \sqrt {d+e x}}{12 b e^2}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e}-\frac {\left ((b d-a e)^2 (5 b B d-6 A b e+a B e)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{16 b e^3}\\ &=\frac {(b d-a e) (5 b B d-6 A b e+a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b e^3}-\frac {(5 b B d-6 A b e+a B e) (a+b x)^{3/2} \sqrt {d+e x}}{12 b e^2}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e}-\frac {\left ((b d-a e)^2 (5 b B d-6 A b e+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 )}{8 b^2 e^3}\\ &=\frac {(b d-a e) (5 b B d-6 A b e+a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b e^3}-\frac {(5 b B d-6 A b e+a B e) (a+b x)^{3/2} \sqrt {d+e x}}{12 b e^2}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e}-\frac {\left ((b d-a e)^2 (5 b B d-6 A b e+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 )}{8 b^2 e^3}\\ &=\frac {(b d-a e) (5 b B d-6 A b e+a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b e^3}-\frac {(5 b B d-6 A b e+a B e) (a+b x)^{3/2} \sqrt {d+e x}}{12 b e^2}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 b e}-\frac {(b d-a e)^2 (5 b B d-6 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{3/2} e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.14, size = 201, normalized size = 1.04 \[ \frac {\sqrt {d+e x} \left (8 B e^3 (a+b x)^3-\frac {(a B e-6 A b e+5 b B d) \left (e (a+b x) \sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}} (5 a e-3 b d+2 b e x)+3 \sqrt {e} \sqrt {a+b x} (b d-a e)^2 \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )\right )}{\sqrt {b d-a e} \sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{24 b e^4 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(8*B*e^3*(a + b*x)^3 - ((5*b*B*d - 6*A*b*e + a*B*e)*(e*Sqrt[b*d - a*e]*(a + b*x)*Sqrt[(b*(d + e
*x))/(b*d - a*e)]*(-3*b*d + 5*a*e + 2*b*e*x) + 3*Sqrt[e]*(b*d - a*e)^2*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)])))/(24*b*e^4*Sqrt[a + b*x])

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fricas [A]  time = 1.03, size = 540, normalized size = 2.80 \[ \left [-\frac {3 \, {\left (5 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} d^{2} e + 3 \, {\left (B a^{2} b + 4 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (8 \, B b^{3} e^{3} x^{2} + 15 \, B b^{3} d^{2} e - 2 \, {\left (11 \, B a b^{2} + 9 \, A b^{3}\right )} d e^{2} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} e^{3} - 2 \, {\left (5 \, B b^{3} d e^{2} - {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{96 \, b^{2} e^{4}}, \frac {3 \, {\left (5 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} d^{2} e + 3 \, {\left (B a^{2} b + 4 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, B b^{3} e^{3} x^{2} + 15 \, B b^{3} d^{2} e - 2 \, {\left (11 \, B a b^{2} + 9 \, A b^{3}\right )} d e^{2} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} e^{3} - 2 \, {\left (5 \, B b^{3} d e^{2} - {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{48 \, b^{2} e^{4}}\right ] \]

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]

[-1/96*(3*(5*B*b^3*d^3 - 3*(3*B*a*b^2 + 2*A*b^3)*d^2*e + 3*(B*a^2*b + 4*A*a*b^2)*d*e^2 + (B*a^3 - 6*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 + 15*B*b^3*d^2*e - 2*(11*B*a*b^2 + 9*A*b^3)*
d*e^2 + 3*(B*a^2*b + 10*A*a*b^2)*e^3 - 2*(5*B*b^3*d*e^2 - (7*B*a*b^2 + 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x
+ d))/(b^2*e^4), 1/48*(3*(5*B*b^3*d^3 - 3*(3*B*a*b^2 + 2*A*b^3)*d^2*e + 3*(B*a^2*b + 4*A*a*b^2)*d*e^2 + (B*a^
3 - 6*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 + 15*B*b^3*d^2*e - 2*(11*B*a*b^2 + 9*A*b^3)*d*e
^2 + 3*(B*a^2*b + 10*A*a*b^2)*e^3 - 2*(5*B*b^3*d*e^2 - (7*B*a*b^2 + 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x +
d))/(b^2*e^4)]

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giac [A]  time = 1.49, size = 268, normalized size = 1.39 \[ \frac {{\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B e^{\left (-1\right )}}{b^{2}} - \frac {{\left (5 \, B b^{3} d e^{3} + B a b^{2} e^{4} - 6 \, A b^{3} e^{4}\right )} e^{\left (-5\right )}}{b^{4}}\right )} + \frac {3 \, {\left (5 \, B b^{4} d^{2} e^{2} - 4 \, B a b^{3} d e^{3} - 6 \, A b^{4} d e^{3} - B a^{2} b^{2} e^{4} + 6 \, A a b^{3} e^{4}\right )} e^{\left (-5\right )}}{b^{4}}\right )} + \frac {3 \, {\left (5 \, B b^{3} d^{3} - 9 \, B a b^{2} d^{2} e - 6 \, A b^{3} d^{2} e + 3 \, B a^{2} b d e^{2} + 12 \, A a b^{2} d e^{2} + B a^{3} e^{3} - 6 \, A a^{2} b e^{3}\right )} e^{\left (-\frac {7}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac {3}{2}}}\right )} b}{24 \, {\left | b \right |}} \]

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="giac")

[Out]

1/24*(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*e^(-1)/b^2 - (5*B*b^3*d*e^
3 + B*a*b^2*e^4 - 6*A*b^3*e^4)*e^(-5)/b^4) + 3*(5*B*b^4*d^2*e^2 - 4*B*a*b^3*d*e^3 - 6*A*b^4*d*e^3 - B*a^2*b^2*
e^4 + 6*A*a*b^3*e^4)*e^(-5)/b^4) + 3*(5*B*b^3*d^3 - 9*B*a*b^2*d^2*e - 6*A*b^3*d^2*e + 3*B*a^2*b*d*e^2 + 12*A*a
*b^2*d*e^2 + B*a^3*e^3 - 6*A*a^2*b*e^3)*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^(3/2))*b/abs(b)

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maple [B]  time = 0.02, size = 636, normalized size = 3.30 \[ \frac {\sqrt {b x +a}\, \sqrt {e x +d}\, \left (18 A \,a^{2} b \,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 )-36 A a \,b^{2} 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 )+18 A \,b^{3} 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 )-3 B \,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 )-9 B \,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 )+27 B 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 )-15 B \,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 {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B \,b^{2} e^{2} x^{2}+24 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,b^{2} e^{2} x +28 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b \,e^{2} x -20 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d e x +60 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, A a b \,e^{2}-36 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, A \,b^{2} d e +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,a^{2} e^{2}-44 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B a b d e +30 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B \,b^{2} d^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b \,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]

1/48*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(16*B*x^2*b^2*e^2*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)+18*A*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*e^3-36*A*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*e^2+18*A*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^2*e+24*A*((b*x+a)*(e*x+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*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*a^3*e^3-9*B*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+27*B*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-15*B*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+28*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*a*b*e^2-20*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*b^2*d*e+60*
A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*e^2-36*A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b^2*d*e+6*B*((b*x+a)*(e
*x+d))^(1/2)*(b*e)^(1/2)*a^2*e^2-44*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*d*e+30*B*(b*e)^(1/2)*((b*x+a)*(e
*x+d))^(1/2)*b^2*d^2)/b/e^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 \[ \text {Exception raised: ValueError} \]

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 = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{\sqrt {d+e\,x}} \,d x \]

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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