3.23.14 \(\int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {d+e x}} \, dx\) [2214]

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

[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 = 0.09, 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 (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} \end {gather*}

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 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)^{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 ) \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^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 ) \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^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 = 0.44, size = 165, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {d+e x} \left (3 a^2 B e^2+2 a b e (-11 B d+15 A e+7 B e x)+b^2 \left (6 A e (-3 d+2 e x)+B \left (15 d^2-10 d e x+8 e^2 x^2\right )\right )\right )}{24 b e^3}-\frac {(b d-a e)^2 (5 b B d-6 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{8 b^{3/2} e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[d + e*x]*(3*a^2*B*e^2 + 2*a*b*e*(-11*B*d + 15*A*e + 7*B*e*x) + b^2*(6*A*e*(-3*d + 2*e*x) +
B*(15*d^2 - 10*d*e*x + 8*e^2*x^2))))/(24*b*e^3) - ((b*d - a*e)^2*(5*b*B*d - 6*A*b*e + a*B*e)*ArcTanh[(Sqrt[b]
*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/(8*b^(3/2)*e^(7/2))

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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 = 0.09, size = 636, normalized size = 3.30

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(e*x+d)^(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-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))*a^3*e^3-9*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^2*b*d*e^2+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*b^2*d^2*e-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))*b^3
*d^3+28*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*e^2*x-20*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b^2*d*e*x+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*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+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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 1.39, size = 520, normalized size = 2.69 \begin {gather*} \left [-\frac {{\left (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^{\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 (15 \, B b^{3} d^{2} e + {\left (8 \, B b^{3} x^{2} + 3 \, B a^{2} b + 30 \, A a b^{2} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x\right )} e^{3} - 2 \, {\left (5 \, B b^{3} d x + {\left (11 \, B a b^{2} + 9 \, A b^{3}\right )} d\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}\right )} e^{\left (-4\right )}}{96 \, b^{2}}, \frac {{\left (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 (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 (15 \, B b^{3} d^{2} e + {\left (8 \, B b^{3} x^{2} + 3 \, B a^{2} b + 30 \, A a b^{2} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x\right )} e^{3} - 2 \, {\left (5 \, B b^{3} d x + {\left (11 \, B a b^{2} + 9 \, A b^{3}\right )} d\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d}\right )} e^{\left (-4\right )}}{48 \, b^{2}}\right ] \end {gather*}

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^(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*(15*B*b^3*d^2*e + (8*B*b^3*x^2 + 3*B*a^2*b + 30*A*a
*b^2 + 2*(7*B*a*b^2 + 6*A*b^3)*x)*e^3 - 2*(5*B*b^3*d*x + (11*B*a*b^2 + 9*A*b^3)*d)*e^2)*sqrt(b*x + a)*sqrt(x*e
+ d))*e^(-4)/b^2, 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*(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*(15*B*b^3*d^2*e + (8*B*b^3*x^2 + 3*B*a^2*b + 30*A*a*b^2 + 2*(7*B*
a*b^2 + 6*A*b^3)*x)*e^3 - 2*(5*B*b^3*d*x + (11*B*a*b^2 + 9*A*b^3)*d)*e^2)*sqrt(b*x + a)*sqrt(x*e + d))*e^(-4)/
b^2]

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

[Out]

Timed out

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Giac [A]
time = 1.31, size = 268, normalized size = 1.39 \begin {gather*} \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 |}} \end {gather*}

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

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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