\(\int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx\) [1810]

Optimal result

Integrand size = 33, antiderivative size = 174 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {(2 b B d+3 A b e-5 a B e) \sqrt {d+e x}}{b^3}+\frac {(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}-\frac {\sqrt {b d-a e} (2 b B d+3 A b e-5 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 79, 52, 65, 214} \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {\sqrt {b d-a e} (-5 a B e+3 A b e+2 b B d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {\sqrt {d+e x} (-5 a B e+3 A b e+2 b B d)}{b^3}+\frac {(d+e x)^{3/2} (-5 a B e+3 A b e+2 b B d)}{3 b^2 (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {d+e x} \left (3 A b (-b d+3 a e+2 b e x)+B \left (-15 a^2 e+a b (11 d-10 e x)+2 b^2 x (4 d+e x)\right )\right )}{3 b^3 (a+b x)}-\frac {\sqrt {-b d+a e} (2 b B d+3 A b e-5 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{7/2}} \]

[In]

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

[Out]

(Sqrt[d + e*x]*(3*A*b*(-(b*d) + 3*a*e + 2*b*e*x) + B*(-15*a^2*e + a*b*(11*d - 10*e*x) + 2*b^2*x*(4*d + e*x))))
/(3*b^3*(a + b*x)) - (Sqrt[-(b*d) + a*e]*(2*b*B*d + 3*A*b*e - 5*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b
*d) + a*e]])/b^(7/2)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {2 \left (B b e x +3 A b e -6 B a e +4 B b d \right ) \sqrt {e x +d}}{3 b^{3}}-\frac {\left (2 a e -2 b d \right ) \left (\frac {\left (-\frac {1}{2} A b e +\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (3 A b e -5 B a e +2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{3}}\)	\(138\)
pseudoelliptic	\(\frac {-3 \left (a e -b d \right ) \left (\left (A e +\frac {2 B d}{3}\right ) b -\frac {5 B a e}{3}\right ) \left (b x +a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )+3 \left (\frac {\left (2 \left (\frac {B x}{3}+A \right ) x e -d \left (-\frac {8 B x}{3}+A \right )\right ) b^{2}}{3}+\left (e \left (-\frac {10 B x}{9}+A \right )+\frac {11 B d}{9}\right ) a b -\frac {5 B e \,a^{2}}{3}\right ) \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}}{b^{3} \left (b x +a \right ) \sqrt {\left (a e -b d \right ) b}}\)	\(148\)
derivativedivides	\(\frac {\frac {2 B \left (e x +d \right )^{\frac {3}{2}} b}{3}+2 A b e \sqrt {e x +d}-4 B a e \sqrt {e x +d}+2 B b d \sqrt {e x +d}}{b^{3}}-\frac {2 \left (\frac {\left (-\frac {1}{2} A a b \,e^{2}+\frac {1}{2} A \,b^{2} d e +\frac {1}{2} a^{2} B \,e^{2}-\frac {1}{2} B a b d e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (3 A a b \,e^{2}-3 A \,b^{2} d e -5 a^{2} B \,e^{2}+7 B a b d e -2 B \,b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{3}}\)	\(195\)
default	\(\frac {\frac {2 B \left (e x +d \right )^{\frac {3}{2}} b}{3}+2 A b e \sqrt {e x +d}-4 B a e \sqrt {e x +d}+2 B b d \sqrt {e x +d}}{b^{3}}-\frac {2 \left (\frac {\left (-\frac {1}{2} A a b \,e^{2}+\frac {1}{2} A \,b^{2} d e +\frac {1}{2} a^{2} B \,e^{2}-\frac {1}{2} B a b d e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (3 A a b \,e^{2}-3 A \,b^{2} d e -5 a^{2} B \,e^{2}+7 B a b d e -2 B \,b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{3}}\)	\(195\)

[In]

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

[Out]

2/3*(B*b*e*x+3*A*b*e-6*B*a*e+4*B*b*d)*(e*x+d)^(1/2)/b^3-1/b^3*(2*a*e-2*b*d)*((-1/2*A*b*e+1/2*B*a*e)*(e*x+d)^(1
/2)/(b*(e*x+d)+a*e-b*d)+1/2*(3*A*b*e-5*B*a*e+2*B*b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)
^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left [\frac {3 \, {\left (2 \, B a b d - {\left (5 \, B a^{2} - 3 \, A a b\right )} e + {\left (2 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, B b^{2} e x^{2} + {\left (11 \, B a b - 3 \, A b^{2}\right )} d - 3 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} e + 2 \, {\left (4 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {3 \, {\left (2 \, B a b d - {\left (5 \, B a^{2} - 3 \, A a b\right )} e + {\left (2 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (2 \, B b^{2} e x^{2} + {\left (11 \, B a b - 3 \, A b^{2}\right )} d - 3 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} e + 2 \, {\left (4 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/6*(3*(2*B*a*b*d - (5*B*a^2 - 3*A*a*b)*e + (2*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt((b*d - a*e)/b)*log((b
*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(2*B*b^2*e*x^2 + (11*B*a*b - 3*A*b^
2)*d - 3*(5*B*a^2 - 3*A*a*b)*e + 2*(4*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(e*x + d))/(b^4*x + a*b^3), -1/3
*(3*(2*B*a*b*d - (5*B*a^2 - 3*A*a*b)*e + (2*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(-(b*d - a*e)/b)*arctan(-s
qrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*B*b^2*e*x^2 + (11*B*a*b - 3*A*b^2)*d - 3*(5*B*a^2 - 3*A*
a*b)*e + 2*(4*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(e*x + d))/(b^4*x + a*b^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.29 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (2 \, B b^{2} d^{2} - 7 \, B a b d e + 3 \, A b^{2} d e + 5 \, B a^{2} e^{2} - 3 \, A a b e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {\sqrt {e x + d} B a b d e - \sqrt {e x + d} A b^{2} d e - \sqrt {e x + d} B a^{2} e^{2} + \sqrt {e x + d} A a b e^{2}}{{\left ({\left (e x + d\right )} b - b d + a e\right )} b^{3}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b^{4} + 3 \, \sqrt {e x + d} B b^{4} d - 6 \, \sqrt {e x + d} B a b^{3} e + 3 \, \sqrt {e x + d} A b^{4} e\right )}}{3 \, b^{6}} \]

[In]

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

[Out]

(2*B*b^2*d^2 - 7*B*a*b*d*e + 3*A*b^2*d*e + 5*B*a^2*e^2 - 3*A*a*b*e^2)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b
*e))/(sqrt(-b^2*d + a*b*e)*b^3) + (sqrt(e*x + d)*B*a*b*d*e - sqrt(e*x + d)*A*b^2*d*e - sqrt(e*x + d)*B*a^2*e^2
+ sqrt(e*x + d)*A*a*b*e^2)/(((e*x + d)*b - b*d + a*e)*b^3) + 2/3*((e*x + d)^(3/2)*B*b^4 + 3*sqrt(e*x + d)*B*b
^4*d - 6*sqrt(e*x + d)*B*a*b^3*e + 3*sqrt(e*x + d)*A*b^4*e)/b^6

Mupad [B] (verification not implemented)

Time = 10.66 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left (\frac {2\,A\,e-2\,B\,d}{b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{b^4}\right )\,\sqrt {d+e\,x}-\frac {\sqrt {d+e\,x}\,\left (B\,a^2\,e^2-A\,a\,b\,e^2-B\,d\,a\,b\,e+A\,d\,b^2\,e\right )}{b^4\,\left (d+e\,x\right )-b^4\,d+a\,b^3\,e}+\frac {2\,B\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,1{}\mathrm {i}}{\sqrt {b\,d-a\,e}}\right )\,\sqrt {b\,d-a\,e}\,\left (3\,A\,b\,e-5\,B\,a\,e+2\,B\,b\,d\right )\,1{}\mathrm {i}}{b^{7/2}} \]

[In]

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

[Out]

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