3.682 \(\int \frac {x^2 (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=309 \[ -\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{64 b d^5}+\frac {5 (a+b x)^{3/2} \sqrt {c+d x} \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{96 b d^4}+\frac {5 (b c-a d)^2 \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{11/2}}+\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (\frac {a^2 d}{b}+14 a c-\frac {63 b c^2}{d}\right )}{24 d^2 (b c-a d)}+\frac {2 c^2 (a+b x)^{7/2}}{d^2 \sqrt {c+d x} (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2} \]

[Out]

5/64*(-a*d+b*c)^2*(-a^2*d^2-14*a*b*c*d+63*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2
)/d^(11/2)+2*c^2*(b*x+a)^(7/2)/d^2/(-a*d+b*c)/(d*x+c)^(1/2)+5/96*(-a^2*d^2-14*a*b*c*d+63*b^2*c^2)*(b*x+a)^(3/2
)*(d*x+c)^(1/2)/b/d^4+1/24*(14*a*c-63*b*c^2/d+a^2*d/b)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d^2/(-a*d+b*c)+1/4*(b*x+a)^
(7/2)*(d*x+c)^(1/2)/b/d^2-5/64*(-a*d+b*c)*(-a^2*d^2-14*a*b*c*d+63*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^5

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Rubi [A]  time = 0.33, antiderivative size = 309, 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, 80, 50, 63, 217, 206} \[ \frac {5 (a+b x)^{3/2} \sqrt {c+d x} \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{96 b d^4}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right )}{64 b d^5}+\frac {5 (b c-a d)^2 \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{11/2}}+\frac {(a+b x)^{5/2} \sqrt {c+d x} \left (\frac {a^2 d}{b}+14 a c-\frac {63 b c^2}{d}\right )}{24 d^2 (b c-a d)}+\frac {2 c^2 (a+b x)^{7/2}}{d^2 \sqrt {c+d x} (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^(5/2))/(c + d*x)^(3/2),x]

[Out]

(2*c^2*(a + b*x)^(7/2))/(d^2*(b*c - a*d)*Sqrt[c + d*x]) - (5*(b*c - a*d)*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*S
qrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^5) + (5*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])
/(96*b*d^4) + ((14*a*c - (63*b*c^2)/d + (a^2*d)/b)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*d^2*(b*c - a*d)) + ((a +
b*x)^(7/2)*Sqrt[c + d*x])/(4*b*d^2) + (5*(b*c - a*d)^2*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*S
qrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(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 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 (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {2 \int \frac {(a+b x)^{5/2} \left (\frac {1}{2} c (7 b c-a d)-\frac {1}{2} d (b c-a d) x\right )}{\sqrt {c+d x}} \, dx}{d^2 (b c-a d)}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{8 b d^2 (b c-a d)}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^3 (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}+\frac {\left (5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right )\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{48 b d^3}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}+\frac {5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^4}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^3 (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}-\frac {\left (5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right )\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{64 b d^4}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^5}+\frac {5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^4}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^3 (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}+\frac {\left (5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b d^5}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^5}+\frac {5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^4}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^3 (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}+\frac {\left (5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-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 )}{64 b^2 d^5}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^5}+\frac {5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^4}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^3 (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}+\frac {\left (5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-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 )}{64 b^2 d^5}\\ &=\frac {2 c^2 (a+b x)^{7/2}}{d^2 (b c-a d) \sqrt {c+d x}}-\frac {5 (b c-a d) \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^5}+\frac {5 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^4}-\frac {\left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^3 (b c-a d)}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d^2}+\frac {5 (b c-a d)^2 \left (63 b^2 c^2-14 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 296, normalized size = 0.96 \[ \frac {15 \left (-a^2 d^2-14 a b c d+63 b^2 c^2\right ) (b c-a d)^{5/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )+\frac {b \sqrt {d} \left (15 a^4 d^3 (c+d x)+a^3 b d^2 \left (-839 c^2-322 c d x+133 d^2 x^2\right )+a^2 b^2 d \left (1785 c^3-202 c^2 d x-581 c d^2 x^2+254 d^3 x^3\right )+a b^3 \left (-945 c^4+1470 c^3 d x+763 c^2 d^2 x^2-316 c d^3 x^3+184 d^4 x^4\right )+3 b^4 x \left (-315 c^4-105 c^3 d x+42 c^2 d^2 x^2-24 c d^3 x^3+16 d^4 x^4\right )\right )}{\sqrt {a+b x}}}{192 b^2 d^{11/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x)^(5/2))/(c + d*x)^(3/2),x]

[Out]

((b*Sqrt[d]*(15*a^4*d^3*(c + d*x) + a^3*b*d^2*(-839*c^2 - 322*c*d*x + 133*d^2*x^2) + a^2*b^2*d*(1785*c^3 - 202
*c^2*d*x - 581*c*d^2*x^2 + 254*d^3*x^3) + 3*b^4*x*(-315*c^4 - 105*c^3*d*x + 42*c^2*d^2*x^2 - 24*c*d^3*x^3 + 16
*d^4*x^4) + a*b^3*(-945*c^4 + 1470*c^3*d*x + 763*c^2*d^2*x^2 - 316*c*d^3*x^3 + 184*d^4*x^4)))/Sqrt[a + b*x] +
15*(b*c - a*d)^(5/2)*(63*b^2*c^2 - 14*a*b*c*d - a^2*d^2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt
[a + b*x])/Sqrt[b*c - a*d]])/(192*b^2*d^(11/2)*Sqrt[c + d*x])

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fricas [A]  time = 2.21, size = 796, normalized size = 2.58 \[ \left [-\frac {15 \, {\left (63 \, b^{4} c^{5} - 140 \, a b^{3} c^{4} d + 90 \, a^{2} b^{2} c^{3} d^{2} - 12 \, a^{3} b c^{2} d^{3} - a^{4} c d^{4} + {\left (63 \, b^{4} c^{4} d - 140 \, a b^{3} c^{3} d^{2} + 90 \, a^{2} b^{2} c^{2} d^{3} - 12 \, a^{3} b c d^{4} - a^{4} d^{5}\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 (48 \, b^{4} d^{5} x^{4} - 945 \, b^{4} c^{4} d + 1785 \, a b^{3} c^{3} d^{2} - 839 \, a^{2} b^{2} c^{2} d^{3} + 15 \, a^{3} b c d^{4} - 8 \, {\left (9 \, b^{4} c d^{4} - 17 \, a b^{3} d^{5}\right )} x^{3} + 2 \, {\left (63 \, b^{4} c^{2} d^{3} - 122 \, a b^{3} c d^{4} + 59 \, a^{2} b^{2} d^{5}\right )} x^{2} - {\left (315 \, b^{4} c^{3} d^{2} - 637 \, a b^{3} c^{2} d^{3} + 337 \, a^{2} b^{2} c d^{4} - 15 \, a^{3} b d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, {\left (b^{2} d^{7} x + b^{2} c d^{6}\right )}}, -\frac {15 \, {\left (63 \, b^{4} c^{5} - 140 \, a b^{3} c^{4} d + 90 \, a^{2} b^{2} c^{3} d^{2} - 12 \, a^{3} b c^{2} d^{3} - a^{4} c d^{4} + {\left (63 \, b^{4} c^{4} d - 140 \, a b^{3} c^{3} d^{2} + 90 \, a^{2} b^{2} c^{2} d^{3} - 12 \, a^{3} b c d^{4} - a^{4} d^{5}\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 (48 \, b^{4} d^{5} x^{4} - 945 \, b^{4} c^{4} d + 1785 \, a b^{3} c^{3} d^{2} - 839 \, a^{2} b^{2} c^{2} d^{3} + 15 \, a^{3} b c d^{4} - 8 \, {\left (9 \, b^{4} c d^{4} - 17 \, a b^{3} d^{5}\right )} x^{3} + 2 \, {\left (63 \, b^{4} c^{2} d^{3} - 122 \, a b^{3} c d^{4} + 59 \, a^{2} b^{2} d^{5}\right )} x^{2} - {\left (315 \, b^{4} c^{3} d^{2} - 637 \, a b^{3} c^{2} d^{3} + 337 \, a^{2} b^{2} c d^{4} - 15 \, a^{3} b d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, {\left (b^{2} d^{7} x + b^{2} c d^{6}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/768*(15*(63*b^4*c^5 - 140*a*b^3*c^4*d + 90*a^2*b^2*c^3*d^2 - 12*a^3*b*c^2*d^3 - a^4*c*d^4 + (63*b^4*c^4*d
- 140*a*b^3*c^3*d^2 + 90*a^2*b^2*c^2*d^3 - 12*a^3*b*c*d^4 - a^4*d^5)*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*(48*b^4*d^5*x^4 - 945*b^4*c^4*d + 1785*a*b^3*c^3*d^2 - 839*a^2*b^2*c^2*d^3 + 15*a^3*b*c*d^4 - 8*(9*b^4*
c*d^4 - 17*a*b^3*d^5)*x^3 + 2*(63*b^4*c^2*d^3 - 122*a*b^3*c*d^4 + 59*a^2*b^2*d^5)*x^2 - (315*b^4*c^3*d^2 - 637
*a*b^3*c^2*d^3 + 337*a^2*b^2*c*d^4 - 15*a^3*b*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^7*x + b^2*c*d^6), -1
/384*(15*(63*b^4*c^5 - 140*a*b^3*c^4*d + 90*a^2*b^2*c^3*d^2 - 12*a^3*b*c^2*d^3 - a^4*c*d^4 + (63*b^4*c^4*d - 1
40*a*b^3*c^3*d^2 + 90*a^2*b^2*c^2*d^3 - 12*a^3*b*c*d^4 - a^4*d^5)*x)*sqrt(-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*(48*b^4*d^5*x^4
- 945*b^4*c^4*d + 1785*a*b^3*c^3*d^2 - 839*a^2*b^2*c^2*d^3 + 15*a^3*b*c*d^4 - 8*(9*b^4*c*d^4 - 17*a*b^3*d^5)*
x^3 + 2*(63*b^4*c^2*d^3 - 122*a*b^3*c*d^4 + 59*a^2*b^2*d^5)*x^2 - (315*b^4*c^3*d^2 - 637*a*b^3*c^2*d^3 + 337*a
^2*b^2*c*d^4 - 15*a^3*b*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^7*x + b^2*c*d^6)]

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giac [A]  time = 1.44, size = 378, normalized size = 1.22 \[ \frac {{\left ({\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{d {\left | b \right |}} - \frac {9 \, b^{3} c d^{7} + 7 \, a b^{2} d^{8}}{b^{2} d^{9} {\left | b \right |}}\right )} + \frac {63 \, b^{4} c^{2} d^{6} - 14 \, a b^{3} c d^{7} - a^{2} b^{2} d^{8}}{b^{2} d^{9} {\left | b \right |}}\right )} - \frac {5 \, {\left (63 \, b^{5} c^{3} d^{5} - 77 \, a b^{4} c^{2} d^{6} + 13 \, a^{2} b^{3} c d^{7} + a^{3} b^{2} d^{8}\right )}}{b^{2} d^{9} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (63 \, b^{6} c^{4} d^{4} - 140 \, a b^{5} c^{3} d^{5} + 90 \, a^{2} b^{4} c^{2} d^{6} - 12 \, a^{3} b^{3} c d^{7} - a^{4} b^{2} d^{8}\right )}}{b^{2} d^{9} {\left | b \right |}}\right )} \sqrt {b x + a}}{192 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {5 \, {\left (63 \, b^{4} c^{4} - 140 \, a b^{3} c^{3} d + 90 \, a^{2} b^{2} c^{2} d^{2} - 12 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{64 \, \sqrt {b d} d^{5} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

1/192*((2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/(d*abs(b)) - (9*b^3*c*d^7 + 7*a*b^2*d^8)/(b^2*d^9*abs(b))) + (63
*b^4*c^2*d^6 - 14*a*b^3*c*d^7 - a^2*b^2*d^8)/(b^2*d^9*abs(b))) - 5*(63*b^5*c^3*d^5 - 77*a*b^4*c^2*d^6 + 13*a^2
*b^3*c*d^7 + a^3*b^2*d^8)/(b^2*d^9*abs(b)))*(b*x + a) - 15*(63*b^6*c^4*d^4 - 140*a*b^5*c^3*d^5 + 90*a^2*b^4*c^
2*d^6 - 12*a^3*b^3*c*d^7 - a^4*b^2*d^8)/(b^2*d^9*abs(b)))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) -
5/64*(63*b^4*c^4 - 140*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 - 12*a^3*b*c*d^3 - a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*
x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^5*abs(b))

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maple [B]  time = 0.03, size = 961, normalized size = 3.11 \[ -\frac {\sqrt {b x +a}\, \left (15 a^{4} d^{5} 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 )+180 a^{3} b c \,d^{4} 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 )-1350 a^{2} b^{2} c^{2} 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 )+2100 a \,b^{3} c^{3} 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 )-945 b^{4} c^{4} 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 )-96 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} d^{4} x^{4}+15 a^{4} c \,d^{4} \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 )+180 a^{3} b \,c^{2} 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 )-1350 a^{2} b^{2} c^{3} 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 )+2100 a \,b^{3} c^{4} 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 )-272 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} d^{4} x^{3}-945 b^{4} c^{5} \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 )+144 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c \,d^{3} x^{3}-236 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b \,d^{4} x^{2}+488 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c \,d^{3} x^{2}-252 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c^{2} d^{2} x^{2}-30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} d^{4} x +674 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b c \,d^{3} x -1274 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c^{2} d^{2} x +630 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c^{3} d x -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} c \,d^{3}+1678 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b \,c^{2} d^{2}-3570 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c^{3} d +1890 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c^{4}\right )}{384 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {d x +c}\, b \,d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^(5/2)/(d*x+c)^(3/2),x)

[Out]

-1/384*(b*x+a)^(1/2)*(-96*x^4*b^3*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-272*x^3*a*b^2*d^4*((b*x+a)*(d*x+c))^
(1/2)*(b*d)^(1/2)+144*x^3*b^3*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+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*a^4*d^5+180*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*c*d^4-1350*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^2*c^2*d^3+2100*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
^3*c^3*d^2-945*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*b^4*c^4*d-236*x^2
*a^2*b*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+488*x^2*a*b^2*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-252*x^2
*b^3*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+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^4*c*d^4+180*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*c^2*d^3-1350*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^2*c^3*d^2+2
100*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*b^3*c^4*d-945*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))*b^4*c^5-30*x*a^3*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*
d)^(1/2)+674*x*a^2*b*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-1274*x*a*b^2*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b
*d)^(1/2)+630*x*b^3*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-30*a^3*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)
+1678*a^2*b*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-3570*a*b^2*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+1
890*b^3*c^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(d*x+c)^(1/2)/b/d^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^(5/2)/(d*x+c)^(3/2),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 \[ \int \frac {x^2\,{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*x)^(5/2))/(c + d*x)^(3/2),x)

[Out]

int((x^2*(a + b*x)^(5/2))/(c + d*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**(5/2)/(d*x+c)**(3/2),x)

[Out]

Timed out

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