\(\int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx\) [627]

Optimal result

Integrand size = 20, antiderivative size = 171 \[ \int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^3}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}-\frac {(b c-a d)^2 (5 b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{7/2}} \]

[Out]

-1/8*(-a*d+b*c)^2*(a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2)/d^(7/2)-1/12*(a*d+5
*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b/d^2+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b/d+1/8*(-a*d+b*c)*(a*d+5*b*c)*(b*x+a)
^(1/2)*(d*x+c)^(1/2)/b/d^3

Rubi [A] (verified)

Time = 0.06 (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 = {81, 52, 65, 223, 212} \[ \int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=-\frac {(b c-a d)^2 (a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{7/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d) (a d+5 b c)}{8 b d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d} \]

[In]

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

[Out]

((b*c - a*d)*(5*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b*d^3) - ((5*b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*
x])/(12*b*d^2) + ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*b*d) - ((b*c - a*d)^2*(5*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[
a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(3/2)*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 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*} \text {integral}& = \frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}-\frac {(5 b c+a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{6 b d} \\ & = -\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}+\frac {((b c-a d) (5 b c+a d)) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 b d^2} \\ & = \frac {(b c-a d) (5 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^3}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b d^3} \\ & = \frac {(b c-a d) (5 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^3}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{8 b^2 d^3} \\ & = \frac {(b c-a d) (5 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^3}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 b^2 d^3} \\ & = \frac {(b c-a d) (5 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b d^3}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b d}-\frac {(b c-a d)^2 (5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.80 \[ \int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^2 d^2+2 a b d (-11 c+7 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 b d^3}-\frac {(b c-a d)^2 (5 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{3/2} d^{7/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(139)=278\).

Time = 0.55 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.31

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}-27 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d +15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}-28 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x +20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+44 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 b \,d^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}\)	\(395\)

[In]

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

[Out]

-1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(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^3*d^3+9*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^2-27*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*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^3*c^3
-28*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c*d*x-6*(b*d)^(1/
2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+44*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d-30*(b*d)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)*b^2*c^2)/b/d^3/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.41 \[ \int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{2} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{2} d^{4}}\right ] \]

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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*(8
*b^3*d^3*x^2 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3)*x)*sqrt(b*x + a)*sq
rt(d*x + c))/(b^2*d^4), 1/48*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d^3*x^2 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^2*d^4)]

Sympy [F]

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

[In]

integrate(x*(b*x+a)**(3/2)/(d*x+c)**(1/2),x)

[Out]

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

Maxima [F(-2)]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.25 \[ \int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2} d} - \frac {5 \, b^{3} c d^{3} + a b^{2} d^{4}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} - a^{2} b^{2} d^{4}\right )}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{3}}\right )} b}{24 \, {\left | b \right |}} \]

[In]

integrate(x*(b*x+a)^(3/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(b^2*d) - (5*b^3*c*d^3 + a*b
^2*d^4)/(b^4*d^5)) + 3*(5*b^4*c^2*d^2 - 4*a*b^3*c*d^3 - a^2*b^2*d^4)/(b^4*d^5)) + 3*(5*b^3*c^3 - 9*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 + a^3*d^3)*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))*b/abs(b)

Mupad [F(-1)]

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

[In]

int((x*(a + b*x)^(3/2))/(c + d*x)^(1/2),x)

[Out]

int((x*(a + b*x)^(3/2))/(c + d*x)^(1/2), x)