\(\int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx\) [469]

Optimal result

Integrand size = 20, antiderivative size = 204 \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=-\frac {a (4 b c-9 a d) (b c-a d) \sqrt {c+d x}}{b^5}-\frac {a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac {a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}+\frac {a (4 b c-9 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {91, 81, 52, 65, 214} \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=-\frac {a^2 (c+d x)^{7/2}}{b^2 (a+b x) (b c-a d)}+\frac {a (4 b c-9 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}}-\frac {a \sqrt {c+d x} (4 b c-9 a d) (b c-a d)}{b^5}-\frac {a (c+d x)^{3/2} (4 b c-9 a d)}{3 b^4}-\frac {a (c+d x)^{5/2} (4 b c-9 a d)}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d} \]

[In]

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

[Out]

-((a*(4*b*c - 9*a*d)*(b*c - a*d)*Sqrt[c + d*x])/b^5) - (a*(4*b*c - 9*a*d)*(c + d*x)^(3/2))/(3*b^4) - (a*(4*b*c
- 9*a*d)*(c + d*x)^(5/2))/(5*b^3*(b*c - a*d)) + (2*(c + d*x)^(7/2))/(7*b^2*d) - (a^2*(c + d*x)^(7/2))/(b^2*(b
*c - a*d)*(a + b*x)) + (a*(4*b*c - 9*a*d)*(b*c - a*d)^(3/2)*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 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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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}& = -\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}+\frac {\int \frac {(c+d x)^{5/2} \left (-\frac {1}{2} a (2 b c-7 a d)+b (b c-a d) x\right )}{a+b x} \, dx}{b^2 (b c-a d)} \\ & = \frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac {(a (4 b c-9 a d)) \int \frac {(c+d x)^{5/2}}{a+b x} \, dx}{2 b^2 (b c-a d)} \\ & = -\frac {a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac {(a (4 b c-9 a d)) \int \frac {(c+d x)^{3/2}}{a+b x} \, dx}{2 b^3} \\ & = -\frac {a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac {a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac {(a (4 b c-9 a d) (b c-a d)) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{2 b^4} \\ & = -\frac {a (4 b c-9 a d) (b c-a d) \sqrt {c+d x}}{b^5}-\frac {a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac {a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac {\left (a (4 b c-9 a d) (b c-a d)^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 b^5} \\ & = -\frac {a (4 b c-9 a d) (b c-a d) \sqrt {c+d x}}{b^5}-\frac {a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac {a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac {\left (a (4 b c-9 a d) (b c-a d)^2\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 {a (4 b c-9 a d) (b c-a d) \sqrt {c+d x}}{b^5}-\frac {a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac {a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d}-\frac {a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}+\frac {a (4 b c-9 a d) (b c-a d)^{3/2} \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 = 0.41 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=\frac {\sqrt {c+d x} \left (-945 a^4 d^3+210 a^3 b d^2 (8 c-3 d x)+30 b^4 x (c+d x)^3+7 a^2 b^2 d \left (-107 c^2+166 c d x+18 d^2 x^2\right )+2 a b^3 \left (15 c^3-277 c^2 d x-109 c d^2 x^2-27 d^3 x^3\right )\right )}{105 b^5 d (a+b x)}-\frac {a (4 b c-9 a d) (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{11/2}} \]

[In]

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

[Out]

(Sqrt[c + d*x]*(-945*a^4*d^3 + 210*a^3*b*d^2*(8*c - 3*d*x) + 30*b^4*x*(c + d*x)^3 + 7*a^2*b^2*d*(-107*c^2 + 16
6*c*d*x + 18*d^2*x^2) + 2*a*b^3*(15*c^3 - 277*c^2*d*x - 109*c*d^2*x^2 - 27*d^3*x^3)))/(105*b^5*d*(a + b*x)) -
(a*(4*b*c - 9*a*d)*(-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(11/2)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {9 \left (a d -b c \right )^{2} \left (a d -\frac {4 b c}{9}\right ) d \left (b x +a \right ) a \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-9 \sqrt {d x +c}\, \sqrt {\left (a d -b c \right ) b}\, \left (-\frac {2 x \left (d x +c \right )^{3} b^{4}}{63}-\frac {2 \left (-\frac {9}{5} d^{3} x^{3}-\frac {109}{15} c \,d^{2} x^{2}-\frac {277}{15} c^{2} d x +c^{3}\right ) a \,b^{3}}{63}+\frac {107 \left (-\frac {18}{107} d^{2} x^{2}-\frac {166}{107} c d x +c^{2}\right ) d \,a^{2} b^{2}}{135}-\frac {16 \left (-\frac {3 d x}{8}+c \right ) d^{2} a^{3} b}{9}+a^{4} d^{3}\right )}{b^{5} d \left (b x +a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(194\)
risch	\(-\frac {2 \left (-15 d^{3} x^{3} b^{3}+42 x^{2} a \,b^{2} d^{3}-45 x^{2} b^{3} c \,d^{2}-105 x \,a^{2} b \,d^{3}+154 x a \,b^{2} c \,d^{2}-45 x \,b^{3} c^{2} d +420 a^{3} d^{3}-735 a^{2} b c \,d^{2}+322 a \,b^{2} c^{2} d -15 b^{3} c^{3}\right ) \sqrt {d x +c}}{105 d \,b^{5}}+\frac {a \left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (-\frac {a d \sqrt {d x +c}}{2 \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\left (9 a d -4 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{5}}\)	\(222\)
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (d x +c \right )^{\frac {7}{2}} b^{3}}{7}+\frac {2 a d \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-a^{2} b \,d^{2} \left (d x +c \right )^{\frac {3}{2}}+\frac {2 a \,b^{2} c d \left (d x +c \right )^{\frac {3}{2}}}{3}+4 \sqrt {d x +c}\, a^{3} d^{3}-6 \sqrt {d x +c}\, a^{2} b c \,d^{2}+2 \sqrt {d x +c}\, a \,b^{2} c^{2} d \right )}{b^{5}}+\frac {2 d a \left (\frac {\left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (9 a^{3} d^{3}-22 a^{2} b c \,d^{2}+17 a \,b^{2} c^{2} d -4 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{5}}}{d}\)	\(252\)
default	\(\frac {-\frac {2 \left (-\frac {\left (d x +c \right )^{\frac {7}{2}} b^{3}}{7}+\frac {2 a d \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}-a^{2} b \,d^{2} \left (d x +c \right )^{\frac {3}{2}}+\frac {2 a \,b^{2} c d \left (d x +c \right )^{\frac {3}{2}}}{3}+4 \sqrt {d x +c}\, a^{3} d^{3}-6 \sqrt {d x +c}\, a^{2} b c \,d^{2}+2 \sqrt {d x +c}\, a \,b^{2} c^{2} d \right )}{b^{5}}+\frac {2 d a \left (\frac {\left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (9 a^{3} d^{3}-22 a^{2} b c \,d^{2}+17 a \,b^{2} c^{2} d -4 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{5}}}{d}\)	\(252\)

[In]

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

[Out]

9*((a*d-b*c)^2*(a*d-4/9*b*c)*d*(b*x+a)*a*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-(d*x+c)^(1/2)*((a*d-b*c)*
b)^(1/2)*(-2/63*x*(d*x+c)^3*b^4-2/63*(-9/5*d^3*x^3-109/15*c*d^2*x^2-277/15*c^2*d*x+c^3)*a*b^3+107/135*(-18/107
*d^2*x^2-166/107*c*d*x+c^2)*d*a^2*b^2-16/9*(-3/8*d*x+c)*d^2*a^3*b+a^4*d^3))/((a*d-b*c)*b)^(1/2)/d/b^5/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.98 \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=\left [\frac {105 \, {\left (4 \, a^{2} b^{2} c^{2} d - 13 \, a^{3} b c d^{2} + 9 \, a^{4} d^{3} + {\left (4 \, a b^{3} c^{2} d - 13 \, a^{2} b^{2} c d^{2} + 9 \, a^{3} b d^{3}\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (30 \, b^{4} d^{3} x^{4} + 30 \, a b^{3} c^{3} - 749 \, a^{2} b^{2} c^{2} d + 1680 \, a^{3} b c d^{2} - 945 \, a^{4} d^{3} + 18 \, {\left (5 \, b^{4} c d^{2} - 3 \, a b^{3} d^{3}\right )} x^{3} + 2 \, {\left (45 \, b^{4} c^{2} d - 109 \, a b^{3} c d^{2} + 63 \, a^{2} b^{2} d^{3}\right )} x^{2} + 2 \, {\left (15 \, b^{4} c^{3} - 277 \, a b^{3} c^{2} d + 581 \, a^{2} b^{2} c d^{2} - 315 \, a^{3} b d^{3}\right )} x\right )} \sqrt {d x + c}}{210 \, {\left (b^{6} d x + a b^{5} d\right )}}, \frac {105 \, {\left (4 \, a^{2} b^{2} c^{2} d - 13 \, a^{3} b c d^{2} + 9 \, a^{4} d^{3} + {\left (4 \, a b^{3} c^{2} d - 13 \, a^{2} b^{2} c d^{2} + 9 \, a^{3} b d^{3}\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (30 \, b^{4} d^{3} x^{4} + 30 \, a b^{3} c^{3} - 749 \, a^{2} b^{2} c^{2} d + 1680 \, a^{3} b c d^{2} - 945 \, a^{4} d^{3} + 18 \, {\left (5 \, b^{4} c d^{2} - 3 \, a b^{3} d^{3}\right )} x^{3} + 2 \, {\left (45 \, b^{4} c^{2} d - 109 \, a b^{3} c d^{2} + 63 \, a^{2} b^{2} d^{3}\right )} x^{2} + 2 \, {\left (15 \, b^{4} c^{3} - 277 \, a b^{3} c^{2} d + 581 \, a^{2} b^{2} c d^{2} - 315 \, a^{3} b d^{3}\right )} x\right )} \sqrt {d x + c}}{105 \, {\left (b^{6} d x + a b^{5} d\right )}}\right ] \]

[In]

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

[Out]

[1/210*(105*(4*a^2*b^2*c^2*d - 13*a^3*b*c*d^2 + 9*a^4*d^3 + (4*a*b^3*c^2*d - 13*a^2*b^2*c*d^2 + 9*a^3*b*d^3)*x
)*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*(30*b^4
*d^3*x^4 + 30*a*b^3*c^3 - 749*a^2*b^2*c^2*d + 1680*a^3*b*c*d^2 - 945*a^4*d^3 + 18*(5*b^4*c*d^2 - 3*a*b^3*d^3)*
x^3 + 2*(45*b^4*c^2*d - 109*a*b^3*c*d^2 + 63*a^2*b^2*d^3)*x^2 + 2*(15*b^4*c^3 - 277*a*b^3*c^2*d + 581*a^2*b^2*
c*d^2 - 315*a^3*b*d^3)*x)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d), 1/105*(105*(4*a^2*b^2*c^2*d - 13*a^3*b*c*d^2 + 9
*a^4*d^3 + (4*a*b^3*c^2*d - 13*a^2*b^2*c*d^2 + 9*a^3*b*d^3)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sq
rt(-(b*c - a*d)/b)/(b*c - a*d)) + (30*b^4*d^3*x^4 + 30*a*b^3*c^3 - 749*a^2*b^2*c^2*d + 1680*a^3*b*c*d^2 - 945*
a^4*d^3 + 18*(5*b^4*c*d^2 - 3*a*b^3*d^3)*x^3 + 2*(45*b^4*c^2*d - 109*a*b^3*c*d^2 + 63*a^2*b^2*d^3)*x^2 + 2*(15
*b^4*c^3 - 277*a*b^3*c^2*d + 581*a^2*b^2*c*d^2 - 315*a^3*b*d^3)*x)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate(x^2*(d*x+c)^(5/2)/(b*x+a)^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 = 0.27 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.40 \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=-\frac {{\left (4 \, a b^{3} c^{3} - 17 \, a^{2} b^{2} c^{2} d + 22 \, a^{3} b c d^{2} - 9 \, a^{4} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{5}} - \frac {\sqrt {d x + c} a^{2} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a^{3} b c d^{2} + \sqrt {d x + c} a^{4} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{5}} + \frac {2 \, {\left (15 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{12} d^{6} - 42 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{11} d^{7} - 70 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{11} c d^{7} - 210 \, \sqrt {d x + c} a b^{11} c^{2} d^{7} + 105 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{10} d^{8} + 630 \, \sqrt {d x + c} a^{2} b^{10} c d^{8} - 420 \, \sqrt {d x + c} a^{3} b^{9} d^{9}\right )}}{105 \, b^{14} d^{7}} \]

[In]

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

[Out]

-(4*a*b^3*c^3 - 17*a^2*b^2*c^2*d + 22*a^3*b*c*d^2 - 9*a^4*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(s
qrt(-b^2*c + a*b*d)*b^5) - (sqrt(d*x + c)*a^2*b^2*c^2*d - 2*sqrt(d*x + c)*a^3*b*c*d^2 + sqrt(d*x + c)*a^4*d^3)
/(((d*x + c)*b - b*c + a*d)*b^5) + 2/105*(15*(d*x + c)^(7/2)*b^12*d^6 - 42*(d*x + c)^(5/2)*a*b^11*d^7 - 70*(d*
x + c)^(3/2)*a*b^11*c*d^7 - 210*sqrt(d*x + c)*a*b^11*c^2*d^7 + 105*(d*x + c)^(3/2)*a^2*b^10*d^8 + 630*sqrt(d*x
+ c)*a^2*b^10*c*d^8 - 420*sqrt(d*x + c)*a^3*b^9*d^9)/(b^14*d^7)

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.05 \[ \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx=\left (\frac {\left (\frac {4\,c}{b^2\,d}+\frac {4\,\left (a\,d-b\,c\right )}{b^3\,d}\right )\,{\left (a\,d-b\,c\right )}^2}{b^2}-\frac {2\,\left (a\,d-b\,c\right )\,\left (\frac {2\,c^2}{b^2\,d}-\frac {2\,{\left (a\,d-b\,c\right )}^2}{b^4\,d}+\frac {2\,\left (\frac {4\,c}{b^2\,d}+\frac {4\,\left (a\,d-b\,c\right )}{b^3\,d}\right )\,\left (a\,d-b\,c\right )}{b}\right )}{b}\right )\,\sqrt {c+d\,x}+{\left (c+d\,x\right )}^{3/2}\,\left (\frac {2\,c^2}{3\,b^2\,d}-\frac {2\,{\left (a\,d-b\,c\right )}^2}{3\,b^4\,d}+\frac {2\,\left (\frac {4\,c}{b^2\,d}+\frac {4\,\left (a\,d-b\,c\right )}{b^3\,d}\right )\,\left (a\,d-b\,c\right )}{3\,b}\right )-\left (\frac {4\,c}{5\,b^2\,d}+\frac {4\,\left (a\,d-b\,c\right )}{5\,b^3\,d}\right )\,{\left (c+d\,x\right )}^{5/2}+\frac {2\,{\left (c+d\,x\right )}^{7/2}}{7\,b^2\,d}-\frac {\sqrt {c+d\,x}\,\left (a^4\,d^3-2\,a^3\,b\,c\,d^2+a^2\,b^2\,c^2\,d\right )}{b^6\,\left (c+d\,x\right )-b^6\,c+a\,b^5\,d}+\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (9\,a\,d-4\,b\,c\right )\,\sqrt {c+d\,x}}{9\,a^4\,d^3-22\,a^3\,b\,c\,d^2+17\,a^2\,b^2\,c^2\,d-4\,a\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (9\,a\,d-4\,b\,c\right )}{b^{11/2}} \]

[In]

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

[Out]

((((4*c)/(b^2*d) + (4*(a*d - b*c))/(b^3*d))*(a*d - b*c)^2)/b^2 - (2*(a*d - b*c)*((2*c^2)/(b^2*d) - (2*(a*d - b
*c)^2)/(b^4*d) + (2*((4*c)/(b^2*d) + (4*(a*d - b*c))/(b^3*d))*(a*d - b*c))/b))/b)*(c + d*x)^(1/2) + (c + d*x)^
(3/2)*((2*c^2)/(3*b^2*d) - (2*(a*d - b*c)^2)/(3*b^4*d) + (2*((4*c)/(b^2*d) + (4*(a*d - b*c))/(b^3*d))*(a*d - b
*c))/(3*b)) - ((4*c)/(5*b^2*d) + (4*(a*d - b*c))/(5*b^3*d))*(c + d*x)^(5/2) + (2*(c + d*x)^(7/2))/(7*b^2*d) -
((c + d*x)^(1/2)*(a^4*d^3 + a^2*b^2*c^2*d - 2*a^3*b*c*d^2))/(b^6*(c + d*x) - b^6*c + a*b^5*d) + (a*atan((a*b^(
1/2)*(a*d - b*c)^(3/2)*(9*a*d - 4*b*c)*(c + d*x)^(1/2))/(9*a^4*d^3 - 4*a*b^3*c^3 + 17*a^2*b^2*c^2*d - 22*a^3*b
*c*d^2))*(a*d - b*c)^(3/2)*(9*a*d - 4*b*c))/b^(11/2)