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

Optimal. Leaf size=204 \[ -\frac {a^2 (c+d x)^{7/2}}{b^2 (a+b x) (b c-a d)}+\frac {a (4 b c-9 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{11/2}}-\frac {a \sqrt {c+d x} (4 b c-9 a d) (b c-a d)}{b^5}-\frac {a (c+d x)^{3/2} (4 b c-9 a d)}{3 b^4}-\frac {a (c+d x)^{5/2} (4 b c-9 a d)}{5 b^3 (b c-a d)}+\frac {2 (c+d x)^{7/2}}{7 b^2 d} \]

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Rubi [A]  time = 0.22, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {89, 80, 50, 63, 208} \begin {gather*} -\frac {a^2 (c+d x)^{7/2}}{b^2 (a+b x) (b c-a d)}-\frac {a (c+d x)^{5/2} (4 b c-9 a d)}{5 b^3 (b c-a d)}-\frac {a (c+d x)^{3/2} (4 b c-9 a d)}{3 b^4}-\frac {a \sqrt {c+d x} (4 b c-9 a d) (b c-a d)}{b^5}+\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}}+\frac {2 (c+d x)^{7/2}}{7 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[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 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 80

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 89

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 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)^2} \, dx &=-\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 ) \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 {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*}

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Mathematica [A]  time = 0.37, size = 181, normalized size = 0.89 \begin {gather*} \frac {\frac {a (9 a d-4 b c) \left (\sqrt {b} \sqrt {c+d x} \left (15 a^2 d^2-5 a b d (7 c+d x)+b^2 \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )-15 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )\right )}{15 b^{7/2}}-\frac {a^2 (c+d x)^{7/2}}{a+b x}+\frac {2 (c+d x)^{7/2} (b c-a d)}{7 d}}{b^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 308, normalized size = 1.51 \begin {gather*} \frac {\sqrt {c+d x} \left (-945 a^4 d^4-630 a^3 b d^3 (c+d x)+2310 a^3 b c d^3-1785 a^2 b^2 c^2 d^2+126 a^2 b^2 d^2 (c+d x)^2+910 a^2 b^2 c d^2 (c+d x)+420 a b^3 c^3 d-280 a b^3 c^2 d (c+d x)-54 a b^3 d (c+d x)^3-56 a b^3 c d (c+d x)^2+30 b^4 (c+d x)^4-30 b^4 c (c+d x)^3\right )}{105 b^5 d (a d+b (c+d x)-b c)}+\frac {\left (-9 a^5 d^4+31 a^4 b c d^3-39 a^3 b^2 c^2 d^2+21 a^2 b^3 c^3 d-4 a b^4 c^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{b^{11/2} (a d-b c)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(420*a*b^3*c^3*d - 1785*a^2*b^2*c^2*d^2 + 2310*a^3*b*c*d^3 - 945*a^4*d^4 - 280*a*b^3*c^2*d*(c +
d*x) + 910*a^2*b^2*c*d^2*(c + d*x) - 630*a^3*b*d^3*(c + d*x) - 56*a*b^3*c*d*(c + d*x)^2 + 126*a^2*b^2*d^2*(c
+ d*x)^2 - 30*b^4*c*(c + d*x)^3 - 54*a*b^3*d*(c + d*x)^3 + 30*b^4*(c + d*x)^4))/(105*b^5*d*(-(b*c) + a*d + b*(
c + d*x))) + ((-4*a*b^4*c^4 + 21*a^2*b^3*c^3*d - 39*a^3*b^2*c^2*d^2 + 31*a^4*b*c*d^3 - 9*a^5*d^4)*ArcTan[(Sqrt
[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/(b^(11/2)*(-(b*c) + a*d)^(3/2))

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fricas [A]  time = 2.20, size = 607, normalized size = 2.98 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 1.31, size = 285, normalized size = 1.40 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.02, size = 377, normalized size = 1.85 \begin {gather*} \frac {9 a^{4} 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 {22 a^{3} 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 {17 a^{2} 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 {4 a \,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 {\sqrt {d x +c}\, a^{4} d^{3}}{\left (b d x +a d \right ) b^{5}}+\frac {2 \sqrt {d x +c}\, a^{3} c \,d^{2}}{\left (b d x +a d \right ) b^{4}}-\frac {\sqrt {d x +c}\, a^{2} c^{2} d}{\left (b d x +a d \right ) b^{3}}-\frac {8 \sqrt {d x +c}\, a^{3} d^{2}}{b^{5}}+\frac {12 \sqrt {d x +c}\, a^{2} c d}{b^{4}}-\frac {4 \sqrt {d x +c}\, a \,c^{2}}{b^{3}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} a^{2} d}{b^{4}}-\frac {4 \left (d x +c \right )^{\frac {3}{2}} a c}{3 b^{3}}-\frac {4 \left (d x +c \right )^{\frac {5}{2}} a}{5 b^{3}}+\frac {2 \left (d x +c \right )^{\frac {7}{2}}}{7 b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(d*x+c)^(7/2)/b^2/d-4/5/b^3*a*(d*x+c)^(5/2)+2*d/b^4*(d*x+c)^(3/2)*a^2-4/3/b^3*(d*x+c)^(3/2)*a*c-8*d^2/b^5*
(d*x+c)^(1/2)*a^3+12*d/b^4*(d*x+c)^(1/2)*a^2*c-4/b^3*(d*x+c)^(1/2)*a*c^2-d^3*a^4/b^5*(d*x+c)^(1/2)/(b*d*x+a*d)
+2*d^2*a^3/b^4*(d*x+c)^(1/2)/(b*d*x+a*d)*c-d*a^2/b^3*(d*x+c)^(1/2)/(b*d*x+a*d)*c^2+9*d^3*a^4/b^5/((a*d-b*c)*b)
^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)-22*d^2*a^3/b^4/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a
*d-b*c)*b)^(1/2)*b)*c+17*d*a^2/b^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*c^2-4*a/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)^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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.46, size = 419, normalized size = 2.05 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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