\(\int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx\) [456]

Optimal result

Integrand size = 20, antiderivative size = 169 \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=\frac {2 a^2 (b c-a d)^2 \sqrt {c+d x}}{b^5}+\frac {2 a^2 (b c-a d) (c+d x)^{3/2}}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}-\frac {2 a^2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}} \]

[Out]

2/3*a^2*(-a*d+b*c)*(d*x+c)^(3/2)/b^4+2/5*a^2*(d*x+c)^(5/2)/b^3-2/7*(a*d+b*c)*(d*x+c)^(7/2)/b^2/d^2+2/9*(d*x+c)
^(9/2)/b/d^2-2*a^2*(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(11/2)+2*a^2*(-a*d+b*c)^
2*(d*x+c)^(1/2)/b^5

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {90, 52, 65, 214} \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=-\frac {2 a^2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}}+\frac {2 a^2 \sqrt {c+d x} (b c-a d)^2}{b^5}+\frac {2 a^2 (c+d x)^{3/2} (b c-a d)}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (c+d x)^{7/2} (a d+b c)}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2} \]

[In]

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

[Out]

(2*a^2*(b*c - a*d)^2*Sqrt[c + d*x])/b^5 + (2*a^2*(b*c - a*d)*(c + d*x)^(3/2))/(3*b^4) + (2*a^2*(c + d*x)^(5/2)
)/(5*b^3) - (2*(b*c + a*d)*(c + d*x)^(7/2))/(7*b^2*d^2) + (2*(c + d*x)^(9/2))/(9*b*d^2) - (2*a^2*(b*c - a*d)^(
5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/b^(11/2)

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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 \left (\frac {(-b c-a d) (c+d x)^{5/2}}{b^2 d}+\frac {a^2 (c+d x)^{5/2}}{b^2 (a+b x)}+\frac {(c+d x)^{7/2}}{b d}\right ) \, dx \\ & = -\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}+\frac {a^2 \int \frac {(c+d x)^{5/2}}{a+b x} \, dx}{b^2} \\ & = \frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}+\frac {\left (a^2 (b c-a d)\right ) \int \frac {(c+d x)^{3/2}}{a+b x} \, dx}{b^3} \\ & = \frac {2 a^2 (b c-a d) (c+d x)^{3/2}}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}+\frac {\left (a^2 (b c-a d)^2\right ) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{b^4} \\ & = \frac {2 a^2 (b c-a d)^2 \sqrt {c+d x}}{b^5}+\frac {2 a^2 (b c-a d) (c+d x)^{3/2}}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}+\frac {\left (a^2 (b c-a d)^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{b^5} \\ & = \frac {2 a^2 (b c-a d)^2 \sqrt {c+d x}}{b^5}+\frac {2 a^2 (b c-a d) (c+d x)^{3/2}}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}+\frac {\left (2 a^2 (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^5 d} \\ & = \frac {2 a^2 (b c-a d)^2 \sqrt {c+d x}}{b^5}+\frac {2 a^2 (b c-a d) (c+d x)^{3/2}}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (b c+a d) (c+d x)^{7/2}}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2}-\frac {2 a^2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=\frac {2 \sqrt {c+d x} \left (315 a^4 d^4-45 a b^3 d (c+d x)^3-5 b^4 (2 c-7 d x) (c+d x)^3-105 a^3 b d^3 (7 c+d x)+21 a^2 b^2 d^2 \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )}{315 b^5 d^2}-\frac {2 a^2 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{11/2}} \]

[In]

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

[Out]

(2*Sqrt[c + d*x]*(315*a^4*d^4 - 45*a*b^3*d*(c + d*x)^3 - 5*b^4*(2*c - 7*d*x)*(c + d*x)^3 - 105*a^3*b*d^3*(7*c
+ d*x) + 21*a^2*b^2*d^2*(23*c^2 + 11*c*d*x + 3*d^2*x^2)))/(315*b^5*d^2) - (2*a^2*(-(b*c) + a*d)^(5/2)*ArcTan[(
Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(11/2)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\sqrt {\left (a d -b c \right ) b}\, \left (-\frac {2 \left (-\frac {7 d x}{2}+c \right ) \left (d x +c \right )^{3} b^{4}}{63}-\frac {a \,b^{3} d \left (d x +c \right )^{3}}{7}+\frac {23 d^{2} \left (\frac {3}{23} d^{2} x^{2}+\frac {11}{23} c d x +c^{2}\right ) a^{2} b^{2}}{15}-\frac {7 \left (\frac {d x}{7}+c \right ) d^{3} a^{3} b}{3}+a^{4} d^{4}\right ) \sqrt {d x +c}+a^{2} d^{2} \left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{\sqrt {\left (a d -b c \right ) b}\, d^{2} b^{5}}\)	\(165\)
derivativedivides	\(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {9}{2}} b^{4}}{9}-\frac {a \,b^{3} d \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {b^{4} c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {a^{2} d^{2} \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {a^{3} b \,d^{3} \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a^{2} b^{2} c \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+a^{4} d^{4} \sqrt {d x +c}-2 a^{3} b c \,d^{3} \sqrt {d x +c}+a^{2} b^{2} c^{2} d^{2} \sqrt {d x +c}\right )}{b^{5}}-\frac {2 a^{2} d^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{5} \sqrt {\left (a d -b c \right ) b}}}{d^{2}}\)	\(236\)
default	\(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {9}{2}} b^{4}}{9}-\frac {a \,b^{3} d \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {b^{4} c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {a^{2} d^{2} \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {a^{3} b \,d^{3} \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a^{2} b^{2} c \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+a^{4} d^{4} \sqrt {d x +c}-2 a^{3} b c \,d^{3} \sqrt {d x +c}+a^{2} b^{2} c^{2} d^{2} \sqrt {d x +c}\right )}{b^{5}}-\frac {2 a^{2} d^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{5} \sqrt {\left (a d -b c \right ) b}}}{d^{2}}\)	\(236\)
risch	\(\frac {2 \left (35 d^{4} x^{4} b^{4}-45 a \,b^{3} d^{4} x^{3}+95 b^{4} c \,d^{3} x^{3}+63 a^{2} b^{2} d^{4} x^{2}-135 a \,b^{3} c \,d^{3} x^{2}+75 b^{4} c^{2} d^{2} x^{2}-105 a^{3} b \,d^{4} x +231 a^{2} b^{2} c \,d^{3} x -135 a \,b^{3} c^{2} d^{2} x +5 b^{4} c^{3} d x +315 a^{4} d^{4}-735 a^{3} b c \,d^{3}+483 a^{2} b^{2} c^{2} d^{2}-45 a \,b^{3} c^{3} d -10 b^{4} c^{4}\right ) \sqrt {d x +c}}{315 d^{2} b^{5}}-\frac {2 a^{2} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{5} \sqrt {\left (a d -b c \right ) b}}\)	\(268\)

[In]

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

[Out]

-2/((a*d-b*c)*b)^(1/2)*(-((a*d-b*c)*b)^(1/2)*(-2/63*(-7/2*d*x+c)*(d*x+c)^3*b^4-1/7*a*b^3*d*(d*x+c)^3+23/15*d^2
*(3/23*d^2*x^2+11/23*c*d*x+c^2)*a^2*b^2-7/3*(1/7*d*x+c)*d^3*a^3*b+a^4*d^4)*(d*x+c)^(1/2)+a^2*d^2*(a*d-b*c)^3*a
rctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))/d^2/b^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 552, normalized size of antiderivative = 3.27 \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=\left [\frac {315 \, {\left (a^{2} b^{2} c^{2} d^{2} - 2 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (35 \, b^{4} d^{4} x^{4} - 10 \, b^{4} c^{4} - 45 \, a b^{3} c^{3} d + 483 \, a^{2} b^{2} c^{2} d^{2} - 735 \, a^{3} b c d^{3} + 315 \, a^{4} d^{4} + 5 \, {\left (19 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{3} + 3 \, {\left (25 \, b^{4} c^{2} d^{2} - 45 \, a b^{3} c d^{3} + 21 \, a^{2} b^{2} d^{4}\right )} x^{2} + {\left (5 \, b^{4} c^{3} d - 135 \, a b^{3} c^{2} d^{2} + 231 \, a^{2} b^{2} c d^{3} - 105 \, a^{3} b d^{4}\right )} x\right )} \sqrt {d x + c}}{315 \, b^{5} d^{2}}, -\frac {2 \, {\left (315 \, {\left (a^{2} b^{2} c^{2} d^{2} - 2 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (35 \, b^{4} d^{4} x^{4} - 10 \, b^{4} c^{4} - 45 \, a b^{3} c^{3} d + 483 \, a^{2} b^{2} c^{2} d^{2} - 735 \, a^{3} b c d^{3} + 315 \, a^{4} d^{4} + 5 \, {\left (19 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{3} + 3 \, {\left (25 \, b^{4} c^{2} d^{2} - 45 \, a b^{3} c d^{3} + 21 \, a^{2} b^{2} d^{4}\right )} x^{2} + {\left (5 \, b^{4} c^{3} d - 135 \, a b^{3} c^{2} d^{2} + 231 \, a^{2} b^{2} c d^{3} - 105 \, a^{3} b d^{4}\right )} x\right )} \sqrt {d x + c}\right )}}{315 \, b^{5} d^{2}}\right ] \]

[In]

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

[Out]

[1/315*(315*(a^2*b^2*c^2*d^2 - 2*a^3*b*c*d^3 + a^4*d^4)*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*(35*b^4*d^4*x^4 - 10*b^4*c^4 - 45*a*b^3*c^3*d + 483*a^2*b^2*c^2
*d^2 - 735*a^3*b*c*d^3 + 315*a^4*d^4 + 5*(19*b^4*c*d^3 - 9*a*b^3*d^4)*x^3 + 3*(25*b^4*c^2*d^2 - 45*a*b^3*c*d^3
+ 21*a^2*b^2*d^4)*x^2 + (5*b^4*c^3*d - 135*a*b^3*c^2*d^2 + 231*a^2*b^2*c*d^3 - 105*a^3*b*d^4)*x)*sqrt(d*x + c
))/(b^5*d^2), -2/315*(315*(a^2*b^2*c^2*d^2 - 2*a^3*b*c*d^3 + a^4*d^4)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x +
c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (35*b^4*d^4*x^4 - 10*b^4*c^4 - 45*a*b^3*c^3*d + 483*a^2*b^2*c^2*d^2 -
735*a^3*b*c*d^3 + 315*a^4*d^4 + 5*(19*b^4*c*d^3 - 9*a*b^3*d^4)*x^3 + 3*(25*b^4*c^2*d^2 - 45*a*b^3*c*d^3 + 21*
a^2*b^2*d^4)*x^2 + (5*b^4*c^3*d - 135*a*b^3*c^2*d^2 + 231*a^2*b^2*c*d^3 - 105*a^3*b*d^4)*x)*sqrt(d*x + c))/(b^
5*d^2)]

Sympy [A] (verification not implemented)

Time = 2.39 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.40 \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {a^{2} d^{3} \left (c + d x\right )^{\frac {5}{2}}}{5 b^{3}} - \frac {a^{2} d^{3} \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{6} \sqrt {\frac {a d - b c}{b}}} + \frac {d \left (c + d x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (c + d x\right )^{\frac {7}{2}} \left (- a d^{2} - b c d\right )}{7 b^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (- a^{3} d^{4} + a^{2} b c d^{3}\right )}{3 b^{4}} + \frac {\sqrt {c + d x} \left (a^{4} d^{5} - 2 a^{3} b c d^{4} + a^{2} b^{2} c^{2} d^{3}\right )}{b^{5}}\right )}{d^{3}} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (\frac {a^{2} \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{2}} - \frac {a x}{b^{2}} + \frac {x^{2}}{2 b}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(a**2*d**3*(c + d*x)**(5/2)/(5*b**3) - a**2*d**3*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*
c)/b))/(b**6*sqrt((a*d - b*c)/b)) + d*(c + d*x)**(9/2)/(9*b) + (c + d*x)**(7/2)*(-a*d**2 - b*c*d)/(7*b**2) + (
c + d*x)**(3/2)*(-a**3*d**4 + a**2*b*c*d**3)/(3*b**4) + sqrt(c + d*x)*(a**4*d**5 - 2*a**3*b*c*d**4 + a**2*b**2
*c**2*d**3)/b**5)/d**3, Ne(d, 0)), (c**(5/2)*(a**2*Piecewise((x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b**2 - a
*x/b**2 + x**2/(2*b)), True))

Maxima [F(-2)]

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

[In]

integrate(x^2*(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 = 0.30 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.49 \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=\frac {2 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{5}} + \frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{8} d^{16} - 45 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{8} c d^{16} - 45 \, {\left (d x + c\right )}^{\frac {7}{2}} a b^{7} d^{17} + 63 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} b^{6} d^{18} + 105 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{6} c d^{18} + 315 \, \sqrt {d x + c} a^{2} b^{6} c^{2} d^{18} - 105 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{3} b^{5} d^{19} - 630 \, \sqrt {d x + c} a^{3} b^{5} c d^{19} + 315 \, \sqrt {d x + c} a^{4} b^{4} d^{20}\right )}}{315 \, b^{9} d^{18}} \]

[In]

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

[Out]

2*(a^2*b^3*c^3 - 3*a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - a^5*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt
(-b^2*c + a*b*d)*b^5) + 2/315*(35*(d*x + c)^(9/2)*b^8*d^16 - 45*(d*x + c)^(7/2)*b^8*c*d^16 - 45*(d*x + c)^(7/2
)*a*b^7*d^17 + 63*(d*x + c)^(5/2)*a^2*b^6*d^18 + 105*(d*x + c)^(3/2)*a^2*b^6*c*d^18 + 315*sqrt(d*x + c)*a^2*b^
6*c^2*d^18 - 105*(d*x + c)^(3/2)*a^3*b^5*d^19 - 630*sqrt(d*x + c)*a^3*b^5*c*d^19 + 315*sqrt(d*x + c)*a^4*b^4*d
^20)/(b^9*d^18)

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 397, normalized size of antiderivative = 2.35 \[ \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx=\left (\frac {2\,c^2}{5\,b\,d^2}+\frac {\left (\frac {4\,c}{b\,d^2}+\frac {2\,\left (a\,d^3-b\,c\,d^2\right )}{b^2\,d^4}\right )\,\left (a\,d^3-b\,c\,d^2\right )}{5\,b\,d^2}\right )\,{\left (c+d\,x\right )}^{5/2}-\left (\frac {4\,c}{7\,b\,d^2}+\frac {2\,\left (a\,d^3-b\,c\,d^2\right )}{7\,b^2\,d^4}\right )\,{\left (c+d\,x\right )}^{7/2}+\frac {2\,{\left (c+d\,x\right )}^{9/2}}{9\,b\,d^2}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {a^2\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}\,\sqrt {c+d\,x}}{a^5\,d^3-3\,a^4\,b\,c\,d^2+3\,a^3\,b^2\,c^2\,d-a^2\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{11/2}}-\frac {\left (\frac {2\,c^2}{b\,d^2}+\frac {\left (\frac {4\,c}{b\,d^2}+\frac {2\,\left (a\,d^3-b\,c\,d^2\right )}{b^2\,d^4}\right )\,\left (a\,d^3-b\,c\,d^2\right )}{b\,d^2}\right )\,\left (a\,d^3-b\,c\,d^2\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,b\,d^2}+\frac {\left (\frac {2\,c^2}{b\,d^2}+\frac {\left (\frac {4\,c}{b\,d^2}+\frac {2\,\left (a\,d^3-b\,c\,d^2\right )}{b^2\,d^4}\right )\,\left (a\,d^3-b\,c\,d^2\right )}{b\,d^2}\right )\,{\left (a\,d^3-b\,c\,d^2\right )}^2\,\sqrt {c+d\,x}}{b^2\,d^4} \]

[In]

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

[Out]

((2*c^2)/(5*b*d^2) + (((4*c)/(b*d^2) + (2*(a*d^3 - b*c*d^2))/(b^2*d^4))*(a*d^3 - b*c*d^2))/(5*b*d^2))*(c + d*x
)^(5/2) - ((4*c)/(7*b*d^2) + (2*(a*d^3 - b*c*d^2))/(7*b^2*d^4))*(c + d*x)^(7/2) + (2*(c + d*x)^(9/2))/(9*b*d^2
) - (2*a^2*atan((a^2*b^(1/2)*(a*d - b*c)^(5/2)*(c + d*x)^(1/2))/(a^5*d^3 - a^2*b^3*c^3 + 3*a^3*b^2*c^2*d - 3*a
^4*b*c*d^2))*(a*d - b*c)^(5/2))/b^(11/2) - (((2*c^2)/(b*d^2) + (((4*c)/(b*d^2) + (2*(a*d^3 - b*c*d^2))/(b^2*d^
4))*(a*d^3 - b*c*d^2))/(b*d^2))*(a*d^3 - b*c*d^2)*(c + d*x)^(3/2))/(3*b*d^2) + (((2*c^2)/(b*d^2) + (((4*c)/(b*
d^2) + (2*(a*d^3 - b*c*d^2))/(b^2*d^4))*(a*d^3 - b*c*d^2))/(b*d^2))*(a*d^3 - b*c*d^2)^2*(c + d*x)^(1/2))/(b^2*
d^4)