\(\int \frac {x (c+d x)^{5/2}}{a+b x} \, dx\) [457]

Optimal result

Integrand size = 18, antiderivative size = 135 \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=-\frac {2 a (b c-a d)^2 \sqrt {c+d x}}{b^4}-\frac {2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d}+\frac {2 a (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}} \]

[Out]

-2/3*a*(-a*d+b*c)*(d*x+c)^(3/2)/b^3-2/5*a*(d*x+c)^(5/2)/b^2+2/7*(d*x+c)^(7/2)/b/d+2*a*(-a*d+b*c)^(5/2)*arctanh
(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(9/2)-2*a*(-a*d+b*c)^2*(d*x+c)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 214} \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=\frac {2 a (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}}-\frac {2 a \sqrt {c+d x} (b c-a d)^2}{b^4}-\frac {2 a (c+d x)^{3/2} (b c-a d)}{3 b^3}-\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d} \]

[In]

Int[(x*(c + d*x)^(5/2))/(a + b*x),x]

[Out]

(-2*a*(b*c - a*d)^2*Sqrt[c + d*x])/b^4 - (2*a*(b*c - a*d)*(c + d*x)^(3/2))/(3*b^3) - (2*a*(c + d*x)^(5/2))/(5*
b^2) + (2*(c + d*x)^(7/2))/(7*b*d) + (2*a*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/
b^(9/2)

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 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}& = \frac {2 (c+d x)^{7/2}}{7 b d}-\frac {a \int \frac {(c+d x)^{5/2}}{a+b x} \, dx}{b} \\ & = -\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d}-\frac {(a (b c-a d)) \int \frac {(c+d x)^{3/2}}{a+b x} \, dx}{b^2} \\ & = -\frac {2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d}-\frac {\left (a (b c-a d)^2\right ) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{b^3} \\ & = -\frac {2 a (b c-a d)^2 \sqrt {c+d x}}{b^4}-\frac {2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d}-\frac {\left (a (b c-a d)^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{b^4} \\ & = -\frac {2 a (b c-a d)^2 \sqrt {c+d x}}{b^4}-\frac {2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d}-\frac {\left (2 a (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{b^4 d} \\ & = -\frac {2 a (b c-a d)^2 \sqrt {c+d x}}{b^4}-\frac {2 a (b c-a d) (c+d x)^{3/2}}{3 b^3}-\frac {2 a (c+d x)^{5/2}}{5 b^2}+\frac {2 (c+d x)^{7/2}}{7 b d}+\frac {2 a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.97 \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=\frac {2 \sqrt {c+d x} \left (-105 a^3 d^3+15 b^3 (c+d x)^3+35 a^2 b d^2 (7 c+d x)-7 a b^2 d \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )}{105 b^4 d}+\frac {2 a (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{9/2}} \]

[In]

Integrate[(x*(c + d*x)^(5/2))/(a + b*x),x]

[Out]

(2*Sqrt[c + d*x]*(-105*a^3*d^3 + 15*b^3*(c + d*x)^3 + 35*a^2*b*d^2*(7*c + d*x) - 7*a*b^2*d*(23*c^2 + 11*c*d*x
+ 3*d^2*x^2)))/(105*b^4*d) + (2*a*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(
9/2)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {\left (d x +c \right )^{3} b^{3}}{7}+\frac {23 d \left (\frac {3}{23} d^{2} x^{2}+\frac {11}{23} c d x +c^{2}\right ) a \,b^{2}}{15}-\frac {7 \left (\frac {d x}{7}+c \right ) d^{2} a^{2} b}{3}+a^{3} d^{3}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}-a d \left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{\sqrt {\left (a d -b c \right ) b}\, d \,b^{4}}\)	\(137\)
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (d x +c \right )^{\frac {7}{2}} b^{3}}{7}+\frac {a d \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {a^{2} b \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a \,b^{2} c d \left (d x +c \right )^{\frac {3}{2}}}{3}+\sqrt {d x +c}\, a^{3} d^{3}-2 \sqrt {d x +c}\, a^{2} b c \,d^{2}+\sqrt {d x +c}\, a \,b^{2} c^{2} d \right )}{b^{4}}+\frac {2 a d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{4} \sqrt {\left (a d -b c \right ) b}}}{d}\)	\(193\)
default	\(\frac {-\frac {2 \left (-\frac {\left (d x +c \right )^{\frac {7}{2}} b^{3}}{7}+\frac {a d \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {a^{2} b \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a \,b^{2} c d \left (d x +c \right )^{\frac {3}{2}}}{3}+\sqrt {d x +c}\, a^{3} d^{3}-2 \sqrt {d x +c}\, a^{2} b c \,d^{2}+\sqrt {d x +c}\, a \,b^{2} c^{2} d \right )}{b^{4}}+\frac {2 a d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{4} \sqrt {\left (a d -b c \right ) b}}}{d}\)	\(193\)
risch	\(-\frac {2 \left (-15 d^{3} x^{3} b^{3}+21 x^{2} a \,b^{2} d^{3}-45 x^{2} b^{3} c \,d^{2}-35 x \,a^{2} b \,d^{3}+77 x a \,b^{2} c \,d^{2}-45 x \,b^{3} c^{2} d +105 a^{3} d^{3}-245 a^{2} b c \,d^{2}+161 a \,b^{2} c^{2} d -15 b^{3} c^{3}\right ) \sqrt {d x +c}}{105 d \,b^{4}}+\frac {2 a \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{4} \sqrt {\left (a d -b c \right ) b}}\)	\(196\)

[In]

int(x*(d*x+c)^(5/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-2/((a*d-b*c)*b)^(1/2)*((-1/7*(d*x+c)^3*b^3+23/15*d*(3/23*d^2*x^2+11/23*c*d*x+c^2)*a*b^2-7/3*(1/7*d*x+c)*d^2*a
^2*b+a^3*d^3)*((a*d-b*c)*b)^(1/2)*(d*x+c)^(1/2)-a*d*(a*d-b*c)^3*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))/d
/b^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 411, normalized size of antiderivative = 3.04 \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=\left [\frac {105 \, {\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (15 \, b^{3} d^{3} x^{3} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \, {\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2} + {\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{105 \, b^{4} d}, \frac {2 \, {\left (105 \, {\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (15 \, b^{3} d^{3} x^{3} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \, {\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2} + {\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}\right )}}{105 \, b^{4} d}\right ] \]

[In]

integrate(x*(d*x+c)^(5/2)/(b*x+a),x, algorithm="fricas")

[Out]

[1/105*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x
+ c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(15*b^3*d^3*x^3 + 15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 -
105*a^3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^2 + (45*b^3*c^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*sqrt(d*x
+ c))/(b^4*d), 2/105*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b
*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (15*b^3*d^3*x^3 + 15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 - 105*a^
3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^2 + (45*b^3*c^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*sqrt(d*x + c))/
(b^4*d)]

Sympy [A] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.40 \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=\begin {cases} \frac {2 \left (- \frac {a d^{2} \left (c + d x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {a d^{2} \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{5} \sqrt {\frac {a d - b c}{b}}} + \frac {d \left (c + d x\right )^{\frac {7}{2}}}{7 b} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (a^{2} d^{3} - a b c d^{2}\right )}{3 b^{3}} + \frac {\sqrt {c + d x} \left (- a^{3} d^{4} + 2 a^{2} b c d^{3} - a b^{2} c^{2} d^{2}\right )}{b^{4}}\right )}{d^{2}} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (- \frac {a \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b} + \frac {x}{b}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x*(d*x+c)**(5/2)/(b*x+a),x)

[Out]

Piecewise((2*(-a*d**2*(c + d*x)**(5/2)/(5*b**2) + a*d**2*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b)
)/(b**5*sqrt((a*d - b*c)/b)) + d*(c + d*x)**(7/2)/(7*b) + (c + d*x)**(3/2)*(a**2*d**3 - a*b*c*d**2)/(3*b**3) +
sqrt(c + d*x)*(-a**3*d**4 + 2*a**2*b*c*d**3 - a*b**2*c**2*d**2)/b**4)/d**2, Ne(d, 0)), (c**(5/2)*(-a*Piecewis
e((x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b + x/b), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x*(d*x+c)^(5/2)/(b*x+a),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 = 0.29 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.57 \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=-\frac {2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{4}} + \frac {2 \, {\left (15 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{6} d^{6} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{5} d^{7} - 35 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{5} c d^{7} - 105 \, \sqrt {d x + c} a b^{5} c^{2} d^{7} + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{4} d^{8} + 210 \, \sqrt {d x + c} a^{2} b^{4} c d^{8} - 105 \, \sqrt {d x + c} a^{3} b^{3} d^{9}\right )}}{105 \, b^{7} d^{7}} \]

[In]

integrate(x*(d*x+c)^(5/2)/(b*x+a),x, algorithm="giac")

[Out]

-2*(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(
-b^2*c + a*b*d)*b^4) + 2/105*(15*(d*x + c)^(7/2)*b^6*d^6 - 21*(d*x + c)^(5/2)*a*b^5*d^7 - 35*(d*x + c)^(3/2)*a
*b^5*c*d^7 - 105*sqrt(d*x + c)*a*b^5*c^2*d^7 + 35*(d*x + c)^(3/2)*a^2*b^4*d^8 + 210*sqrt(d*x + c)*a^2*b^4*c*d^
8 - 105*sqrt(d*x + c)*a^3*b^3*d^9)/(b^7*d^7)

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.82 \[ \int \frac {x (c+d x)^{5/2}}{a+b x} \, dx=\frac {2\,{\left (c+d\,x\right )}^{7/2}}{7\,b\,d}-\left (\frac {2\,c}{5\,b\,d}+\frac {2\,\left (a\,d^2-b\,c\,d\right )}{5\,b^2\,d^2}\right )\,{\left (c+d\,x\right )}^{5/2}+\frac {2\,a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}\,\sqrt {c+d\,x}}{a^4\,d^3-3\,a^3\,b\,c\,d^2+3\,a^2\,b^2\,c^2\,d-a\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{9/2}}-\frac {{\left (a\,d^2-b\,c\,d\right )}^2\,\left (\frac {2\,c}{b\,d}+\frac {2\,\left (a\,d^2-b\,c\,d\right )}{b^2\,d^2}\right )\,\sqrt {c+d\,x}}{b^2\,d^2}+\frac {\left (a\,d^2-b\,c\,d\right )\,\left (\frac {2\,c}{b\,d}+\frac {2\,\left (a\,d^2-b\,c\,d\right )}{b^2\,d^2}\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,b\,d} \]

[In]

int((x*(c + d*x)^(5/2))/(a + b*x),x)

[Out]

(2*(c + d*x)^(7/2))/(7*b*d) - ((2*c)/(5*b*d) + (2*(a*d^2 - b*c*d))/(5*b^2*d^2))*(c + d*x)^(5/2) + (2*a*atan((a
*b^(1/2)*(a*d - b*c)^(5/2)*(c + d*x)^(1/2))/(a^4*d^3 - a*b^3*c^3 + 3*a^2*b^2*c^2*d - 3*a^3*b*c*d^2))*(a*d - b*
c)^(5/2))/b^(9/2) - ((a*d^2 - b*c*d)^2*((2*c)/(b*d) + (2*(a*d^2 - b*c*d))/(b^2*d^2))*(c + d*x)^(1/2))/(b^2*d^2
) + ((a*d^2 - b*c*d)*((2*c)/(b*d) + (2*(a*d^2 - b*c*d))/(b^2*d^2))*(c + d*x)^(3/2))/(3*b*d)