\(\int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\) [761]

Optimal result

Integrand size = 20, antiderivative size = 171 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {3 (b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {(b c-5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2 (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {3 (b c-5 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {3 (b c-5 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} \sqrt {d}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-5 a d)}{4 b^3}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d)}{2 b^2 (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.83 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {\frac {\sqrt {b} \sqrt {c+d x} \left (-15 a^2 d+a b (13 c-5 d x)+b^2 x (5 c+2 d x)\right )}{\sqrt {a+b x}}-\frac {6 \left (b^2 c^2-6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \left (\sqrt {a-\frac {b c}{d}}-\sqrt {a+b x}\right )}\right )}{\sqrt {d}}}{4 b^{7/2}} \]

[In]

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

[Out]

((Sqrt[b]*Sqrt[c + d*x]*(-15*a^2*d + a*b*(13*c - 5*d*x) + b^2*x*(5*c + 2*d*x)))/Sqrt[a + b*x] - (6*(b^2*c^2 -
6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*(Sqrt[a - (b*c)/d] - Sqrt[a + b*x]))])/Sqrt[d]
)/(4*b^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(454\) vs. \(2(141)=282\).

Time = 0.59 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.66

method	result	size
default	\(\frac {\sqrt {d x +c}\, \left (15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{2} x -18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x +4 b^{2} d \,x^{2} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{2}-18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2}-10 a b d x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+10 b^{2} c x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-30 a^{2} d \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+26 a b c \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{3}}\)	\(455\)

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*d^2*x-
18*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c*d*x+3*ln(1/2*(2*b*d*x+2
*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^2*x+4*b^2*d*x^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))
^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*d^2-18*ln(1/2*(2*b*d
*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c*d+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))
^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c^2-10*a*b*d*x*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*b^2*c*x*(
b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*a^2*d*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+26*a*b*c*(b*d)^(1/2)*((b*x+a)*
(d*x+c))^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.54 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{5} d x + a b^{4} d\right )}}, -\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{5} d x + a b^{4} d\right )}}\right ] \]

[In]

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

[Out]

[1/16*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x)*sqrt(b*d)*log(8*b^2*d
^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^
2*c*d + a*b*d^2)*x) + 4*(2*b^3*d^2*x^2 + 13*a*b^2*c*d - 15*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^5*d*x + a*b^4*d), -1/8*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d +
5*a^2*b*d^2)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^
2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 2*(2*b^3*d^2*x^2 + 13*a*b^2*c*d - 15*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^
2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d*x + a*b^4*d)]

Sympy [F]

\[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x \left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x*(d*x+c)**(3/2)/(b*x+a)**(3/2),x)

[Out]

Integral(x*(c + d*x)**(3/2)/(a + b*x)**(3/2), x)

Maxima [F(-2)]

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

[In]

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(3/2),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*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.46 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d {\left | b \right |}}{b^{5}} + \frac {5 \, b^{10} c d^{2} {\left | b \right |} - 9 \, a b^{9} d^{3} {\left | b \right |}}{b^{14} d^{2}}\right )} - \frac {3 \, {\left (b^{2} c^{2} {\left | b \right |} - 6 \, a b c d {\left | b \right |} + 5 \, a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{8 \, \sqrt {b d} b^{4}} + \frac {4 \, {\left (a b^{2} c^{2} d {\left | b \right |} - 2 \, a^{2} b c d^{2} {\left | b \right |} + a^{3} d^{3} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} \sqrt {b d} b^{3}} \]

[In]

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

[Out]

1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d*abs(b)/b^5 + (5*b^10*c*d^2*abs(b) - 9*a*b
^9*d^3*abs(b))/(b^14*d^2)) - 3/8*(b^2*c^2*abs(b) - 6*a*b*c*d*abs(b) + 5*a^2*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*
x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*b^4) + 4*(a*b^2*c^2*d*abs(b) - 2*a^2*b*c*d^2*abs(b
) + a^3*d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*sqrt(
b*d)*b^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x\,{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int((x*(c + d*x)^(3/2))/(a + b*x)^(3/2),x)

[Out]

int((x*(c + d*x)^(3/2))/(a + b*x)^(3/2), x)