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

Optimal. Leaf size=319 \[ -\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (a+b x)^{3/2} (b c-a d)}+\frac {5 (b c-a d) \left (21 a^2 d^2-14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{11/2} \sqrt {d}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{8 b^5}+\frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{12 b^4 (b c-a d)}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{3 b^3 (b c-a d)^2}+\frac {4 a (c+d x)^{7/2} (3 b c-5 a d)}{3 b^2 \sqrt {a+b x} (b c-a d)^2} \]

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Rubi [A]  time = 0.33, antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {89, 78, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x} (c+d x)^{5/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{3 b^3 (b c-a d)^2}+\frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{12 b^4 (b c-a d)}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right )}{8 b^5}+\frac {5 (b c-a d) \left (21 a^2 d^2-14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{11/2} \sqrt {d}}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (a+b x)^{3/2} (b c-a d)}+\frac {4 a (c+d x)^{7/2} (3 b c-5 a d)}{3 b^2 \sqrt {a+b x} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^5) + (5*(b^2*c^2 - 14*a*b*c*d + 21*a^
2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b^4*(b*c - a*d)) + ((b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*Sqrt[a + b*x
]*(c + d*x)^(5/2))/(3*b^3*(b*c - a*d)^2) - (2*a^2*(c + d*x)^(7/2))/(3*b^2*(b*c - a*d)*(a + b*x)^(3/2)) + (4*a*
(3*b*c - 5*a*d)*(c + d*x)^(7/2))/(3*b^2*(b*c - a*d)^2*Sqrt[a + b*x]) + (5*(b*c - a*d)*(b^2*c^2 - 14*a*b*c*d +
21*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(11/2)*Sqrt[d])

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x^2 (c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx &=-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {2 \int \frac {(c+d x)^{5/2} \left (-\frac {1}{2} a (3 b c-7 a d)+\frac {3}{2} b (b c-a d) x\right )}{(a+b x)^{3/2}} \, dx}{3 b^2 (b c-a d)}\\ &=-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx}{b^2 (b c-a d)^2}\\ &=\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{6 b^3 (b c-a d)}\\ &=\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{8 b^4}\\ &=\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^5}+\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (5 (b c-a d) \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b^5}\\ &=\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^5}+\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (5 (b c-a d) \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{8 b^6}\\ &=\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^5}+\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (5 (b c-a d) \left (b^2 c^2-14 a b c d+21 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 b^6}\\ &=\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^5}+\frac {5 \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^4 (b c-a d)}+\frac {\left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^3 (b c-a d)^2}-\frac {2 a^2 (c+d x)^{7/2}}{3 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {4 a (3 b c-5 a d) (c+d x)^{7/2}}{3 b^2 (b c-a d)^2 \sqrt {a+b x}}+\frac {5 (b c-a d) \left (b^2 c^2-14 a b c d+21 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{11/2} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.80, size = 218, normalized size = 0.68 \begin {gather*} \frac {\sqrt {c+d x} \left (\frac {15 \sqrt {b c-a d} \left (21 a^2 d^2-14 a b c d+b^2 c^2\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \sqrt {\frac {b (c+d x)}{b c-a d}}}+\frac {315 a^4 d^2+420 a^3 b d (d x-c)+a^2 b^2 \left (113 c^2-574 c d x+63 d^2 x^2\right )-6 a b^3 x \left (-27 c^2+16 c d x+3 d^2 x^2\right )+b^4 x^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )}{(a+b x)^{3/2}}\right )}{24 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*((315*a^4*d^2 + 420*a^3*b*d*(-c + d*x) - 6*a*b^3*x*(-27*c^2 + 16*c*d*x + 3*d^2*x^2) + b^4*x^2*(
33*c^2 + 26*c*d*x + 8*d^2*x^2) + a^2*b^2*(113*c^2 - 574*c*d*x + 63*d^2*x^2))/(a + b*x)^(3/2) + (15*Sqrt[b*c -
a*d]*(b^2*c^2 - 14*a*b*c*d + 21*a^2*d^2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[(b*(c
+ d*x))/(b*c - a*d)])))/(24*b^5)

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IntegrateAlgebraic [F]  time = 180.02, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 3.29, size = 804, normalized size = 2.52 \begin {gather*} \left [-\frac {15 \, {\left (a^{2} b^{3} c^{3} - 15 \, a^{3} b^{2} c^{2} d + 35 \, a^{4} b c d^{2} - 21 \, a^{5} d^{3} + {\left (b^{5} c^{3} - 15 \, a b^{4} c^{2} d + 35 \, a^{2} b^{3} c d^{2} - 21 \, a^{3} b^{2} d^{3}\right )} x^{2} + 2 \, {\left (a b^{4} c^{3} - 15 \, a^{2} b^{3} c^{2} d + 35 \, a^{3} b^{2} c d^{2} - 21 \, a^{4} b d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{5} d^{3} x^{4} + 113 \, a^{2} b^{3} c^{2} d - 420 \, a^{3} b^{2} c d^{2} + 315 \, a^{4} b d^{3} + 2 \, {\left (13 \, b^{5} c d^{2} - 9 \, a b^{4} d^{3}\right )} x^{3} + 3 \, {\left (11 \, b^{5} c^{2} d - 32 \, a b^{4} c d^{2} + 21 \, a^{2} b^{3} d^{3}\right )} x^{2} + 2 \, {\left (81 \, a b^{4} c^{2} d - 287 \, a^{2} b^{3} c d^{2} + 210 \, a^{3} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{8} d x^{2} + 2 \, a b^{7} d x + a^{2} b^{6} d\right )}}, -\frac {15 \, {\left (a^{2} b^{3} c^{3} - 15 \, a^{3} b^{2} c^{2} d + 35 \, a^{4} b c d^{2} - 21 \, a^{5} d^{3} + {\left (b^{5} c^{3} - 15 \, a b^{4} c^{2} d + 35 \, a^{2} b^{3} c d^{2} - 21 \, a^{3} b^{2} d^{3}\right )} x^{2} + 2 \, {\left (a b^{4} c^{3} - 15 \, a^{2} b^{3} c^{2} d + 35 \, a^{3} b^{2} c d^{2} - 21 \, a^{4} b d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{5} d^{3} x^{4} + 113 \, a^{2} b^{3} c^{2} d - 420 \, a^{3} b^{2} c d^{2} + 315 \, a^{4} b d^{3} + 2 \, {\left (13 \, b^{5} c d^{2} - 9 \, a b^{4} d^{3}\right )} x^{3} + 3 \, {\left (11 \, b^{5} c^{2} d - 32 \, a b^{4} c d^{2} + 21 \, a^{2} b^{3} d^{3}\right )} x^{2} + 2 \, {\left (81 \, a b^{4} c^{2} d - 287 \, a^{2} b^{3} c d^{2} + 210 \, a^{3} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{8} d x^{2} + 2 \, a b^{7} d x + a^{2} b^{6} 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)^(5/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(a^2*b^3*c^3 - 15*a^3*b^2*c^2*d + 35*a^4*b*c*d^2 - 21*a^5*d^3 + (b^5*c^3 - 15*a*b^4*c^2*d + 35*a^2*
b^3*c*d^2 - 21*a^3*b^2*d^3)*x^2 + 2*(a*b^4*c^3 - 15*a^2*b^3*c^2*d + 35*a^3*b^2*c*d^2 - 21*a^4*b*d^3)*x)*sqrt(b
*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d
*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8*b^5*d^3*x^4 + 113*a^2*b^3*c^2*d - 420*a^3*b^2*c*d^2 + 315*a^4*b*d^3
+ 2*(13*b^5*c*d^2 - 9*a*b^4*d^3)*x^3 + 3*(11*b^5*c^2*d - 32*a*b^4*c*d^2 + 21*a^2*b^3*d^3)*x^2 + 2*(81*a*b^4*c^
2*d - 287*a^2*b^3*c*d^2 + 210*a^3*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^8*d*x^2 + 2*a*b^7*d*x + a^2*b^6*
d), -1/48*(15*(a^2*b^3*c^3 - 15*a^3*b^2*c^2*d + 35*a^4*b*c*d^2 - 21*a^5*d^3 + (b^5*c^3 - 15*a*b^4*c^2*d + 35*a
^2*b^3*c*d^2 - 21*a^3*b^2*d^3)*x^2 + 2*(a*b^4*c^3 - 15*a^2*b^3*c^2*d + 35*a^3*b^2*c*d^2 - 21*a^4*b*d^3)*x)*sqr
t(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*
c*d + a*b*d^2)*x)) - 2*(8*b^5*d^3*x^4 + 113*a^2*b^3*c^2*d - 420*a^3*b^2*c*d^2 + 315*a^4*b*d^3 + 2*(13*b^5*c*d^
2 - 9*a*b^4*d^3)*x^3 + 3*(11*b^5*c^2*d - 32*a*b^4*c*d^2 + 21*a^2*b^3*d^3)*x^2 + 2*(81*a*b^4*c^2*d - 287*a^2*b^
3*c*d^2 + 210*a^3*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^8*d*x^2 + 2*a*b^7*d*x + a^2*b^6*d)]

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giac [B]  time = 3.25, size = 927, normalized size = 2.91 \begin {gather*} \frac {1}{24} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{7}} + \frac {13 \, b^{21} c d^{5} {\left | b \right |} - 25 \, a b^{20} d^{6} {\left | b \right |}}{b^{27} d^{4}}\right )} + \frac {3 \, {\left (11 \, b^{22} c^{2} d^{4} {\left | b \right |} - 58 \, a b^{21} c d^{5} {\left | b \right |} + 55 \, a^{2} b^{20} d^{6} {\left | b \right |}\right )}}{b^{27} d^{4}}\right )} - \frac {5 \, {\left (\sqrt {b d} b^{3} c^{3} {\left | b \right |} - 15 \, \sqrt {b d} a b^{2} c^{2} d {\left | b \right |} + 35 \, \sqrt {b d} a^{2} b c d^{2} {\left | b \right |} - 21 \, \sqrt {b d} a^{3} d^{3} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{16 \, b^{7} d} + \frac {4 \, {\left (6 \, \sqrt {b d} a b^{7} c^{5} {\left | b \right |} - 37 \, \sqrt {b d} a^{2} b^{6} c^{4} d {\left | b \right |} + 88 \, \sqrt {b d} a^{3} b^{5} c^{3} d^{2} {\left | b \right |} - 102 \, \sqrt {b d} a^{4} b^{4} c^{2} d^{3} {\left | b \right |} + 58 \, \sqrt {b d} a^{5} b^{3} c d^{4} {\left | b \right |} - 13 \, \sqrt {b d} a^{6} b^{2} d^{5} {\left | b \right |} - 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{5} c^{4} {\left | b \right |} + 60 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{4} c^{3} d {\left | b \right |} - 108 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b^{3} c^{2} d^{2} {\left | b \right |} + 84 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{4} b^{2} c d^{3} {\left | b \right |} - 24 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{5} b d^{4} {\left | b \right |} + 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b^{3} c^{3} {\left | b \right |} - 27 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{2} b^{2} c^{2} d {\left | b \right |} + 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{3} b c d^{2} {\left | b \right |} - 15 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{4} d^{3} {\left | b \right |}\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*d^2*abs(b)/b^7 + (13*b^21*c*d
^5*abs(b) - 25*a*b^20*d^6*abs(b))/(b^27*d^4)) + 3*(11*b^22*c^2*d^4*abs(b) - 58*a*b^21*c*d^5*abs(b) + 55*a^2*b^
20*d^6*abs(b))/(b^27*d^4)) - 5/16*(sqrt(b*d)*b^3*c^3*abs(b) - 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 35*sqrt(b*d)*a
^2*b*c*d^2*abs(b) - 21*sqrt(b*d)*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a
*b*d))^2)/(b^7*d) + 4/3*(6*sqrt(b*d)*a*b^7*c^5*abs(b) - 37*sqrt(b*d)*a^2*b^6*c^4*d*abs(b) + 88*sqrt(b*d)*a^3*b
^5*c^3*d^2*abs(b) - 102*sqrt(b*d)*a^4*b^4*c^2*d^3*abs(b) + 58*sqrt(b*d)*a^5*b^3*c*d^4*abs(b) - 13*sqrt(b*d)*a^
6*b^2*d^5*abs(b) - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^5*c^4*ab
s(b) + 60*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^4*c^3*d*abs(b) - 1
08*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^3*c^2*d^2*abs(b) + 84*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^2*c*d^3*abs(b) - 24*sqrt(b*d)*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b*d^4*abs(b) + 6*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^3*c^3*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^2*c^2*d*abs(b) + 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^
2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b*c*d^2*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d))^4*a^4*d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^2)^3*b^6)

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maple [B]  time = 0.03, size = 1002, normalized size = 3.14 \begin {gather*} -\frac {\sqrt {d x +c}\, \left (315 a^{3} b^{2} d^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-525 a^{2} b^{3} c \,d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+225 a \,b^{4} c^{2} d \,x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 b^{5} c^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+630 a^{4} b \,d^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-1050 a^{3} b^{2} c \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+450 a^{2} b^{3} c^{2} d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-30 a \,b^{4} c^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} d^{2} x^{4}+315 a^{5} d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-525 a^{4} b c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+225 a^{3} b^{2} c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 a^{2} b^{3} c^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+36 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} d^{2} x^{3}-52 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c d \,x^{3}-126 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b^{2} d^{2} x^{2}+192 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{3} c d \,x^{2}-66 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{2} x^{2}-840 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} b \,d^{2} x +1148 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b^{2} c d x -324 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{2} x -630 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{4} d^{2}+840 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} b c d -226 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \left (b x +a \right )^{\frac {3}{2}} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(d*x+c)^(1/2)*(-16*x^4*b^4*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+315*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a
)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^3*b^2*d^3-525*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/
2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^2*b^3*c*d^2+225*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/
2))/(b*d)^(1/2))*x^2*a*b^4*c^2*d-15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2)
)*x^2*b^5*c^3+36*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^3*a*b^3*d^2-52*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^3*
b^4*c*d+630*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a^4*b*d^3-1050*ln(1/
2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a^3*b^2*c*d^2+450*ln(1/2*(2*b*d*x+a*d
+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a^2*b^3*c^2*d-30*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)
*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*a*b^4*c^3-126*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^2*b^2*d^2+
192*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a*b^3*c*d-66*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^2*b^4*c^2+315*l
n(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^5*d^3-525*ln(1/2*(2*b*d*x+a*d+b*c
+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^4*b*c*d^2+225*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c
))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*b^2*c^2*d-15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1
/2))/(b*d)^(1/2))*a^2*b^3*c^3-840*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^3*b*d^2+1148*(b*d)^(1/2)*((b*x+a)*(d
*x+c))^(1/2)*x*a^2*b^2*c*d-324*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*a*b^3*c^2-630*(b*d)^(1/2)*((b*x+a)*(d*x+c
))^(1/2)*a^4*d^2+840*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*c*d-226*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2
*b^2*c^2)/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(3/2)/b^5

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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)^(5/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,{\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**(5/2),x)

[Out]

Timed out

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