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

Optimal. Leaf size=169 \[ -\frac {2 a^2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}}+\frac {2 a^2 \sqrt {c+d x} (b c-a d)^2}{b^5}+\frac {2 a^2 (c+d x)^{3/2} (b c-a d)}{3 b^4}+\frac {2 a^2 (c+d x)^{5/2}}{5 b^3}-\frac {2 (c+d x)^{7/2} (a d+b c)}{7 b^2 d^2}+\frac {2 (c+d x)^{9/2}}{9 b d^2} \]

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Rubi [A]  time = 0.11, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {88, 50, 63, 208} \begin {gather*} \frac {2 a^2 (c+d x)^{5/2}}{5 b^3}+\frac {2 a^2 (c+d x)^{3/2} (b c-a d)}{3 b^4}+\frac {2 a^2 \sqrt {c+d x} (b c-a d)^2}{b^5}-\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}}-\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} \end {gather*}

Antiderivative was successfully verified.

[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 50

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 63

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 88

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {x^2 (c+d x)^{5/2}}{a+b x} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.30, size = 151, normalized size = 0.89 \begin {gather*} \frac {2 \left (\frac {105 a^2 (a d-b c) \left (3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )-\sqrt {b} \sqrt {c+d x} (-3 a d+4 b c+b d x)\right )}{b^{5/2}}+63 a^2 (c+d x)^{5/2}-\frac {45 b (c+d x)^{7/2} (a d+b c)}{d^2}+\frac {35 b^2 (c+d x)^{9/2}}{d^2}\right )}{315 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(63*a^2*(c + d*x)^(5/2) - (45*b*(b*c + a*d)*(c + d*x)^(7/2))/d^2 + (35*b^2*(c + d*x)^(9/2))/d^2 + (105*a^2*
(-(b*c) + a*d)*(-(Sqrt[b]*Sqrt[c + d*x]*(4*b*c - 3*a*d + b*d*x)) + 3*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c
+ d*x])/Sqrt[b*c - a*d]]))/b^(5/2)))/(315*b^3)

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IntegrateAlgebraic [A]  time = 0.22, size = 201, normalized size = 1.19 \begin {gather*} \frac {2 a^2 (a d-b c)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{b^{11/2}}+\frac {2 \sqrt {c+d x} \left (315 a^4 d^4-105 a^3 b d^3 (c+d x)-630 a^3 b c d^3+315 a^2 b^2 c^2 d^2+63 a^2 b^2 d^2 (c+d x)^2+105 a^2 b^2 c d^2 (c+d x)-45 a b^3 d (c+d x)^3+35 b^4 (c+d x)^4-45 b^4 c (c+d x)^3\right )}{315 b^5 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x]*(315*a^2*b^2*c^2*d^2 - 630*a^3*b*c*d^3 + 315*a^4*d^4 + 105*a^2*b^2*c*d^2*(c + d*x) - 105*a^3*
b*d^3*(c + d*x) + 63*a^2*b^2*d^2*(c + d*x)^2 - 45*b^4*c*(c + d*x)^3 - 45*a*b^3*d*(c + d*x)^3 + 35*b^4*(c + d*x
)^4))/(315*b^5*d^2) + (2*a^2*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d
)])/b^(11/2)

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fricas [A]  time = 1.21, size = 552, normalized size = 3.27 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 1.32, size = 252, normalized size = 1.49 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.01, size = 331, normalized size = 1.96 \begin {gather*} -\frac {2 a^{5} d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{5}}+\frac {6 a^{4} c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{4}}-\frac {6 a^{3} c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{3}}+\frac {2 a^{2} c^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}+\frac {2 \sqrt {d x +c}\, a^{4} d^{2}}{b^{5}}-\frac {4 \sqrt {d x +c}\, a^{3} c d}{b^{4}}+\frac {2 \sqrt {d x +c}\, a^{2} c^{2}}{b^{3}}-\frac {2 \left (d x +c \right )^{\frac {3}{2}} a^{3} d}{3 b^{4}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} a^{2} c}{3 b^{3}}+\frac {2 \left (d x +c \right )^{\frac {5}{2}} a^{2}}{5 b^{3}}-\frac {2 \left (d x +c \right )^{\frac {7}{2}} a}{7 b^{2} d}-\frac {2 \left (d x +c \right )^{\frac {7}{2}} c}{7 b \,d^{2}}+\frac {2 \left (d x +c \right )^{\frac {9}{2}}}{9 b \,d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.40, size = 397, normalized size = 2.35 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 57.68, size = 185, normalized size = 1.09 \begin {gather*} \frac {2 a^{2} \left (c + d x\right )^{\frac {5}{2}}}{5 b^{3}} - \frac {2 a^{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^{6} \sqrt {\frac {a d - b c}{b}}} + \frac {2 \left (c + d x\right )^{\frac {9}{2}}}{9 b d^{2}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \left (- 2 a d - 2 b c\right )}{7 b^{2} d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (- 2 a^{3} d + 2 a^{2} b c\right )}{3 b^{4}} + \frac {\sqrt {c + d x} \left (2 a^{4} d^{2} - 4 a^{3} b c d + 2 a^{2} b^{2} c^{2}\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**2*(c + d*x)**(5/2)/(5*b**3) - 2*a**2*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**6*sqrt((a
*d - b*c)/b)) + 2*(c + d*x)**(9/2)/(9*b*d**2) + (c + d*x)**(7/2)*(-2*a*d - 2*b*c)/(7*b**2*d**2) + (c + d*x)**(
3/2)*(-2*a**3*d + 2*a**2*b*c)/(3*b**4) + sqrt(c + d*x)*(2*a**4*d**2 - 4*a**3*b*c*d + 2*a**2*b**2*c**2)/b**5

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