\(\int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx\) [635]

Optimal result

Integrand size = 20, antiderivative size = 174 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}-\frac {3 (5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}+\frac {3 (b c-a d) (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {3 (b c-a d) (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{7/2}}-\frac {3 \sqrt {a+b x} \sqrt {c+d x} (5 b c-a d)}{4 d^3}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 b c-a d)}{2 d^2 (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)} \]

[In]

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

[Out]

(-2*c*(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) - (3*(5*b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^3) +
((5*b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d^2*(b*c - a*d)) + (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(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 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 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}+\frac {(5 b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{d (b c-a d)} \\ & = -\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}-\frac {(3 (5 b c-a d)) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{4 d^2} \\ & = -\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}-\frac {3 (5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}+\frac {(3 (b c-a d) (5 b c-a d)) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 d^3} \\ & = -\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}-\frac {3 (5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}+\frac {(3 (b c-a d) (5 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 d^3} \\ & = -\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}-\frac {3 (5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}+\frac {(3 (b c-a d) (5 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 d^3} \\ & = -\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}-\frac {3 (5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}+\frac {3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.83 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {\frac {\sqrt {d} \sqrt {a+b x} \left (a d (13 c+5 d x)+b \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{\sqrt {c+d x}}-\frac {6 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \left (\sqrt {a-\frac {b c}{d}}-\sqrt {a+b x}\right )}\right )}{\sqrt {b}}}{4 d^{7/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(454\) vs. \(2(144)=288\).

Time = 0.55 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.61

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.49 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + {\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b d^{5} x + b c d^{4}\right )}}, -\frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + {\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b d^{5} x + b c d^{4}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(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 = 0.33 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} {\left | b \right |}}{b d} - \frac {5 \, b^{2} c d^{3} {\left | b \right |} - a b d^{4} {\left | b \right |}}{b^{2} d^{5}}\right )} - \frac {3 \, {\left (5 \, b^{3} c^{2} d^{2} {\left | b \right |} - 6 \, a b^{2} c d^{3} {\left | b \right |} + a^{2} b d^{4} {\left | b \right |}\right )}}{b^{2} d^{5}}\right )}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {3 \, {\left (5 \, b^{2} c^{2} {\left | b \right |} - 6 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} b d^{3}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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