3.23.37 \(\int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx\) [2237]

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

[Out]

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

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Rubi [A]
time = 0.10, 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 = {81, 52, 65, 223, 212} \begin {gather*} -\frac {(b d-a e)^2 (5 a B e-6 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}}-\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (5 a B e-6 A b e+b B d)}{8 b^3 e}-\frac {\sqrt {a+b x} (d+e x)^{3/2} (5 a B e-6 A b e+b B d)}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

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 65

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 81

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx &=\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}+\frac {\left (3 A b e-B \left (\frac {b d}{2}+\frac {5 a e}{2}\right )\right ) \int \frac {(d+e x)^{3/2}}{\sqrt {a+b x}} \, dx}{3 b e}\\ &=-\frac {(b B d-6 A b e+5 a B e) \sqrt {a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}-\frac {((b d-a e) (b B d-6 A b e+5 a B e)) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x}} \, dx}{8 b^2 e}\\ &=-\frac {(b d-a e) (b B d-6 A b e+5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b^3 e}-\frac {(b B d-6 A b e+5 a B e) \sqrt {a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}-\frac {\left ((b d-a e)^2 (b B d-6 A b e+5 a B e)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{16 b^3 e}\\ &=-\frac {(b d-a e) (b B d-6 A b e+5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b^3 e}-\frac {(b B d-6 A b e+5 a B e) \sqrt {a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}-\frac {\left ((b d-a e)^2 (b B d-6 A b e+5 a B e)\right ) \text {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^4 e}\\ &=-\frac {(b d-a e) (b B d-6 A b e+5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b^3 e}-\frac {(b B d-6 A b e+5 a B e) \sqrt {a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}-\frac {\left ((b d-a e)^2 (b B d-6 A b e+5 a B e)\right ) \text {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^4 e}\\ &=-\frac {(b d-a e) (b B d-6 A b e+5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b^3 e}-\frac {(b B d-6 A b e+5 a B e) \sqrt {a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}-\frac {(b d-a e)^2 (b B d-6 A b e+5 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.43, size = 165, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {d+e x} \left (15 a^2 B e^2-2 a b e (11 B d+9 A e+5 B e x)+b^2 \left (6 A e (5 d+2 e x)+B \left (3 d^2+14 d e x+8 e^2 x^2\right )\right )\right )}{24 b^3 e}-\frac {(b d-a e)^2 (b B d-6 A b e+5 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[d + e*x]*(15*a^2*B*e^2 - 2*a*b*e*(11*B*d + 9*A*e + 5*B*e*x) + b^2*(6*A*e*(5*d + 2*e*x) + B
*(3*d^2 + 14*d*e*x + 8*e^2*x^2))))/(24*b^3*e) - ((b*d - a*e)^2*(b*B*d - 6*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sq
rt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(7/2)*e^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(635\) vs. \(2(161)=322\).
time = 0.10, size = 636, normalized size = 3.30

method	result	size
default	\(\frac {\sqrt {e x +d}\, \sqrt {b x +a}\, \left (16 B \,b^{2} e^{2} x^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+18 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} b \,e^{3}-36 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d \,e^{2}+18 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{2} e +24 A \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} e^{2} x -15 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{3} e^{3}+27 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} b d \,e^{2}-9 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d^{2} e -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{3}-20 B \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a b \,e^{2} x +28 B \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} d e x -36 A \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a b \,e^{2}+60 A \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} d e +30 B \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a^{2} e^{2}-44 B \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, a b d e +6 B \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} d^{2}\right )}{48 b^{3} e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}\)	\(636\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(16*B*b^2*e^2*x^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+18*A*ln(1/2*(2*b*e*x+2*
((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*e^3-36*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(
1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d*e^2+18*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+
a*e+b*d)/(b*e)^(1/2))*b^3*d^2*e+24*A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b^2*e^2*x-15*B*ln(1/2*(2*b*e*x+2*((b*
x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*e^3+27*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*d*e^2-9*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d
)/(b*e)^(1/2))*a*b^2*d^2*e-3*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3
*d^3-20*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*e^2*x+28*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b^2*d*e*x-36*
A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*e^2+60*A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b^2*d*e+30*B*(b*e)^(1/2
)*((b*x+a)*(e*x+d))^(1/2)*a^2*e^2-44*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*d*e+6*B*(b*e)^(1/2)*((b*x+a)*(e
*x+d))^(1/2)*b^2*d^2)/b^3/e/((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((B*x+A)*(e*x+d)^(3/2)/(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 1.66, size = 522, normalized size = 2.70 \begin {gather*} \left [-\frac {{\left (3 \, {\left (B b^{3} d^{3} + 3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {b} e^{\frac {1}{2}} \log \left (b^{2} d^{2} + 4 \, {\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\frac {1}{2}} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) - 4 \, {\left (3 \, B b^{3} d^{2} e + {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} e^{3} + 2 \, {\left (7 \, B b^{3} d x - {\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}\right )} e^{\left (-2\right )}}{96 \, b^{4}}, \frac {{\left (3 \, {\left (B b^{3} d^{3} + 3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {-b e} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {-b e} \sqrt {x e + d}}{2 \, {\left ({\left (b^{2} x^{2} + a b x\right )} e^{2} + {\left (b^{2} d x + a b d\right )} e\right )}}\right ) + 2 \, {\left (3 \, B b^{3} d^{2} e + {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} e^{3} + 2 \, {\left (7 \, B b^{3} d x - {\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}\right )} e^{\left (-2\right )}}{48 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a*b^2)*d*e^2 + (5*B*a^3 - 6*A*a^2*b)*e
^3)*sqrt(b)*e^(1/2)*log(b^2*d^2 + 4*(b*d + (2*b*x + a)*e)*sqrt(b*x + a)*sqrt(x*e + d)*sqrt(b)*e^(1/2) + (8*b^2
*x^2 + 8*a*b*x + a^2)*e^2 + 2*(4*b^2*d*x + 3*a*b*d)*e) - 4*(3*B*b^3*d^2*e + (8*B*b^3*x^2 + 15*B*a^2*b - 18*A*a
*b^2 - 2*(5*B*a*b^2 - 6*A*b^3)*x)*e^3 + 2*(7*B*b^3*d*x - (11*B*a*b^2 - 15*A*b^3)*d)*e^2)*sqrt(b*x + a)*sqrt(x*
e + d))*e^(-2)/b^4, 1/48*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a*b^2)*d*e^2 + (5*B*
a^3 - 6*A*a^2*b)*e^3)*sqrt(-b*e)*arctan(1/2*(b*d + (2*b*x + a)*e)*sqrt(b*x + a)*sqrt(-b*e)*sqrt(x*e + d)/((b^2
*x^2 + a*b*x)*e^2 + (b^2*d*x + a*b*d)*e)) + 2*(3*B*b^3*d^2*e + (8*B*b^3*x^2 + 15*B*a^2*b - 18*A*a*b^2 - 2*(5*B
*a*b^2 - 6*A*b^3)*x)*e^3 + 2*(7*B*b^3*d*x - (11*B*a*b^2 - 15*A*b^3)*d)*e^2)*sqrt(b*x + a)*sqrt(x*e + d))*e^(-2
)/b^4]

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Sympy [F(-1)] Timed out
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((B*x+A)*(e*x+d)**(3/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 575 vs. \(2 (168) = 336\).
time = 1.49, size = 575, normalized size = 2.98 \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 d {\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 |} e}{b^{2}} - \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 d {\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 |} e}{b^{3}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(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*d*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)*e/b^2 - 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*d*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)*e/b^3)/b

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{\sqrt {a+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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