3.18.60 \(\int \frac {(A+B x) (d+e x)^{3/2}}{(a+b x)^3} \, dx\) [1760]

Optimal. Leaf size=197 \[ \frac {3 e (4 b B d+A b e-5 a B e) \sqrt {d+e x}}{4 b^3 (b d-a e)}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x)^2}-\frac {3 e (4 b B d+A b e-5 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {b d-a e}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 43, 52, 65, 214} \begin {gather*} -\frac {3 e (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {b d-a e}}+\frac {3 e \sqrt {d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 (b d-a e)}-\frac {(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rubi steps

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

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Mathematica [A]
time = 0.68, size = 139, normalized size = 0.71 \begin {gather*} -\frac {\sqrt {d+e x} \left (A b (2 b d+3 a e+5 b e x)+B \left (-15 a^2 e+a b (2 d-25 e x)+4 b^2 x (d-2 e x)\right )\right )}{4 b^3 (a+b x)^2}+\frac {3 e (4 b B d+A b e-5 a B e) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{7/2} \sqrt {-b d+a e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[d + e*x]*(A*b*(2*b*d + 3*a*e + 5*b*e*x) + B*(-15*a^2*e + a*b*(2*d - 25*e*x) + 4*b^2*x*(d - 2*e*x)))
)/(b^3*(a + b*x)^2) + (3*e*(4*b*B*d + A*b*e - 5*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*
b^(7/2)*Sqrt[-(b*d) + a*e])

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Maple [A]
time = 0.11, size = 171, normalized size = 0.87

method	result	size
derivativedivides	\(2 e \left (\frac {B \sqrt {e x +d}}{b^{3}}+\frac {\frac {\left (-\frac {5}{8} A \,b^{2} e +\frac {9}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {3}{8} A a b \,e^{2}+\frac {3}{8} A \,b^{2} d e +\frac {7}{8} B \,a^{2} e^{2}-\frac {11}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {3 \left (A b e -5 B a e +4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{b^{3}}\right )\)	\(171\)
default	\(2 e \left (\frac {B \sqrt {e x +d}}{b^{3}}+\frac {\frac {\left (-\frac {5}{8} A \,b^{2} e +\frac {9}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {3}{8} A a b \,e^{2}+\frac {3}{8} A \,b^{2} d e +\frac {7}{8} B \,a^{2} e^{2}-\frac {11}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {3 \left (A b e -5 B a e +4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{b^{3}}\right )\)	\(171\)
risch	\(\frac {2 B \sqrt {e x +d}\, e}{b^{3}}-\frac {5 e^{2} \left (e x +d \right )^{\frac {3}{2}} A}{4 b \left (b e x +a e \right )^{2}}+\frac {9 e^{2} \left (e x +d \right )^{\frac {3}{2}} B a}{4 b^{2} \left (b e x +a e \right )^{2}}-\frac {e \left (e x +d \right )^{\frac {3}{2}} B d}{b \left (b e x +a e \right )^{2}}-\frac {3 e^{3} \sqrt {e x +d}\, A a}{4 b^{2} \left (b e x +a e \right )^{2}}+\frac {3 e^{2} \sqrt {e x +d}\, A d}{4 b \left (b e x +a e \right )^{2}}+\frac {7 e^{3} \sqrt {e x +d}\, B \,a^{2}}{4 b^{3} \left (b e x +a e \right )^{2}}-\frac {11 e^{2} \sqrt {e x +d}\, B a d}{4 b^{2} \left (b e x +a e \right )^{2}}+\frac {e \sqrt {e x +d}\, B \,d^{2}}{b \left (b e x +a e \right )^{2}}+\frac {3 e^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) A}{4 b^{2} \sqrt {\left (a e -b d \right ) b}}-\frac {15 e^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) B a}{4 b^{3} \sqrt {\left (a e -b d \right ) b}}+\frac {3 e \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) B d}{b^{2} \sqrt {\left (a e -b d \right ) b}}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*(B/b^3*(e*x+d)^(1/2)+1/b^3*(((-5/8*A*b^2*e+9/8*B*a*b*e-1/2*b^2*B*d)*(e*x+d)^(3/2)+(-3/8*A*a*b*e^2+3/8*A*b^
2*d*e+7/8*B*a^2*e^2-11/8*B*a*b*d*e+1/2*b^2*B*d^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+3/8*(A*b*e-5*B*a*e+4*B*
b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(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)^3,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 = 0.91, size = 679, normalized size = 3.45 \begin {gather*} \left [-\frac {3 \, \sqrt {b^{2} d - a b e} {\left ({\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} e^{2} - 4 \, {\left (B b^{3} d x^{2} + 2 \, B a b^{2} d x + B a^{2} b d\right )} e\right )} \log \left (\frac {2 \, b d + {\left (b x - a\right )} e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {x e + d}}{b x + a}\right ) + 2 \, {\left (4 \, B b^{4} d^{2} x + 2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} + {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} e^{2} - {\left (8 \, B b^{4} d x^{2} + {\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d x + {\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{7} d x^{2} + 2 \, a b^{6} d x + a^{2} b^{5} d - {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )} e\right )}}, -\frac {3 \, \sqrt {-b^{2} d + a b e} {\left ({\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} e^{2} - 4 \, {\left (B b^{3} d x^{2} + 2 \, B a b^{2} d x + B a^{2} b d\right )} e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {x e + d}}{b x e + b d}\right ) + {\left (4 \, B b^{4} d^{2} x + 2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} + {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} e^{2} - {\left (8 \, B b^{4} d x^{2} + {\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d x + {\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{7} d x^{2} + 2 \, a b^{6} d x + a^{2} b^{5} d - {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )} e\right )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/8*(3*sqrt(b^2*d - a*b*e)*((5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a*b^2)*x)*e^2 -
4*(B*b^3*d*x^2 + 2*B*a*b^2*d*x + B*a^2*b*d)*e)*log((2*b*d + (b*x - a)*e - 2*sqrt(b^2*d - a*b*e)*sqrt(x*e + d))
/(b*x + a)) + 2*(4*B*b^4*d^2*x + 2*(B*a*b^3 + A*b^4)*d^2 + (8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*
a^2*b^2 - A*a*b^3)*x)*e^2 - (8*B*b^4*d*x^2 + (29*B*a*b^3 - 5*A*b^4)*d*x + (17*B*a^2*b^2 - A*a*b^3)*d)*e)*sqrt(
x*e + d))/(b^7*d*x^2 + 2*a*b^6*d*x + a^2*b^5*d - (a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4)*e), -1/4*(3*sqrt(-b^2*d +
a*b*e)*((5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a*b^2)*x)*e^2 - 4*(B*b^3*d*x^2 + 2*B*
a*b^2*d*x + B*a^2*b*d)*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(x*e + d)/(b*x*e + b*d)) + (4*B*b^4*d^2*x + 2*(B*a*b
^3 + A*b^4)*d^2 + (8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^2*b^2 - A*a*b^3)*x)*e^2 - (8*B*b^4*d*x^
2 + (29*B*a*b^3 - 5*A*b^4)*d*x + (17*B*a^2*b^2 - A*a*b^3)*d)*e)*sqrt(x*e + d))/(b^7*d*x^2 + 2*a*b^6*d*x + a^2*
b^5*d - (a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4)*e)]

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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)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.45, size = 236, normalized size = 1.20 \begin {gather*} \frac {2 \, \sqrt {x e + d} B e}{b^{3}} + \frac {3 \, {\left (4 \, B b d e - 5 \, B a e^{2} + A b e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{3}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e - 4 \, \sqrt {x e + d} B b^{2} d^{2} e - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{2} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{2} + 11 \, \sqrt {x e + d} B a b d e^{2} - 3 \, \sqrt {x e + d} A b^{2} d e^{2} - 7 \, \sqrt {x e + d} B a^{2} e^{3} + 3 \, \sqrt {x e + d} A a b e^{3}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3}} \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)^3,x, algorithm="giac")

[Out]

2*sqrt(x*e + d)*B*e/b^3 + 3/4*(4*B*b*d*e - 5*B*a*e^2 + A*b*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(
sqrt(-b^2*d + a*b*e)*b^3) - 1/4*(4*(x*e + d)^(3/2)*B*b^2*d*e - 4*sqrt(x*e + d)*B*b^2*d^2*e - 9*(x*e + d)^(3/2)
*B*a*b*e^2 + 5*(x*e + d)^(3/2)*A*b^2*e^2 + 11*sqrt(x*e + d)*B*a*b*d*e^2 - 3*sqrt(x*e + d)*A*b^2*d*e^2 - 7*sqrt
(x*e + d)*B*a^2*e^3 + 3*sqrt(x*e + d)*A*a*b*e^3)/(((x*e + d)*b - b*d + a*e)^2*b^3)

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Mupad [B]
time = 1.30, size = 256, normalized size = 1.30 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {7\,B\,a^2\,e^3}{4}-\frac {11\,B\,a\,b\,d\,e^2}{4}-\frac {3\,A\,a\,b\,e^3}{4}+B\,b^2\,d^2\,e+\frac {3\,A\,b^2\,d\,e^2}{4}\right )-{\left (d+e\,x\right )}^{3/2}\,\left (\frac {5\,A\,b^2\,e^2}{4}+B\,d\,b^2\,e-\frac {9\,B\,a\,b\,e^2}{4}\right )}{b^5\,{\left (d+e\,x\right )}^2-\left (2\,b^5\,d-2\,a\,b^4\,e\right )\,\left (d+e\,x\right )+b^5\,d^2+a^2\,b^3\,e^2-2\,a\,b^4\,d\,e}+\frac {2\,B\,e\,\sqrt {d+e\,x}}{b^3}+\frac {3\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (A\,b\,e-5\,B\,a\,e+4\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^2-5\,B\,a\,e^2+4\,B\,b\,d\,e\right )}\right )\,\left (A\,b\,e-5\,B\,a\,e+4\,B\,b\,d\right )}{4\,b^{7/2}\,\sqrt {a\,e-b\,d}} \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)^3,x)

[Out]

((d + e*x)^(1/2)*((7*B*a^2*e^3)/4 - (3*A*a*b*e^3)/4 + (3*A*b^2*d*e^2)/4 + B*b^2*d^2*e - (11*B*a*b*d*e^2)/4) -
(d + e*x)^(3/2)*((5*A*b^2*e^2)/4 - (9*B*a*b*e^2)/4 + B*b^2*d*e))/(b^5*(d + e*x)^2 - (2*b^5*d - 2*a*b^4*e)*(d +
e*x) + b^5*d^2 + a^2*b^3*e^2 - 2*a*b^4*d*e) + (2*B*e*(d + e*x)^(1/2))/b^3 + (3*e*atan((b^(1/2)*e*(d + e*x)^(1
/2)*(A*b*e - 5*B*a*e + 4*B*b*d))/((a*e - b*d)^(1/2)*(A*b*e^2 - 5*B*a*e^2 + 4*B*b*d*e)))*(A*b*e - 5*B*a*e + 4*B
*b*d))/(4*b^(7/2)*(a*e - b*d)^(1/2))

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