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

Optimal. Leaf size=306 \[ -\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (-\frac {63 a^2 d}{b}+14 a c+\frac {b c^2}{d}\right )}{24 b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 \sqrt {a+b x} (b c-a d)}-\frac {5 (b c-a d)^2 \left (-63 a^2 d^2+14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{11/2} d^{3/2}}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) \left (-63 a^2 d^2+14 a b c d+b^2 c^2\right )}{64 b^5 d}-\frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (-63 a^2 d^2+14 a b c d+b^2 c^2\right )}{96 b^4 d}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d} \]

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Rubi [A]  time = 0.33, antiderivative size = 306, 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} \begin {gather*} -\frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (-63 a^2 d^2+14 a b c d+b^2 c^2\right )}{96 b^4 d}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) \left (-63 a^2 d^2+14 a b c d+b^2 c^2\right )}{64 b^5 d}-\frac {5 (b c-a d)^2 \left (-63 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 )}{64 b^{11/2} d^{3/2}}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (-\frac {63 a^2 d}{b}+14 a c+\frac {b c^2}{d}\right )}{24 b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 \sqrt {a+b x} (b c-a d)}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b*c - a*d)*(b^2*c^2 + 14*a*b*c*d - 63*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^5*d) - (5*(b^2*c^2 + 14
*a*b*c*d - 63*a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(96*b^4*d) - ((14*a*c + (b*c^2)/d - (63*a^2*d)/b)*Sqrt[a
+ b*x]*(c + d*x)^(5/2))/(24*b^2*(b*c - a*d)) - (2*a^2*(c + d*x)^(7/2))/(b^2*(b*c - a*d)*Sqrt[a + b*x]) + (Sqr
t[a + b*x]*(c + d*x)^(7/2))/(4*b^2*d) - (5*(b*c - a*d)^2*(b^2*c^2 + 14*a*b*c*d - 63*a^2*d^2)*ArcTanh[(Sqrt[d]*
Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(11/2)*d^(3/2))

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 (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx &=-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {2 \int \frac {(c+d x)^{5/2} \left (-\frac {1}{2} a (b c-7 a d)+\frac {1}{2} b (b c-a d) x\right )}{\sqrt {a+b x}} \, dx}{b^2 (b c-a d)}\\ &=-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx}{8 b^2 d (b c-a d)}\\ &=-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {\left (5 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right )\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{48 b^3 d}\\ &=-\frac {5 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^4 d}-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {\left (5 (b c-a d) \left (b^2 c^2+14 a b c d-63 a^2 d^2\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{64 b^4 d}\\ &=-\frac {5 (b c-a d) \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^5 d}-\frac {5 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^4 d}-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {\left (5 (b c-a d)^2 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b^5 d}\\ &=-\frac {5 (b c-a d) \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^5 d}-\frac {5 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^4 d}-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {\left (5 (b c-a d)^2 \left (b^2 c^2+14 a b c d-63 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^6 d}\\ &=-\frac {5 (b c-a d) \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^5 d}-\frac {5 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^4 d}-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {\left (5 (b c-a d)^2 \left (b^2 c^2+14 a b c d-63 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^6 d}\\ &=-\frac {5 (b c-a d) \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^5 d}-\frac {5 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^4 d}-\frac {\left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{7/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{7/2}}{4 b^2 d}-\frac {5 (b c-a d)^2 \left (b^2 c^2+14 a b c d-63 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{11/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.91, size = 245, normalized size = 0.80 \begin {gather*} \frac {\sqrt {c+d x} \left (\frac {\sqrt {d} \left (-945 a^4 d^3+105 a^3 b d^2 (17 c-3 d x)+a^2 b^2 d \left (-839 c^2+637 c d x+126 d^2 x^2\right )+a b^3 \left (15 c^3-337 c^2 d x-244 c d^2 x^2-72 d^3 x^3\right )+b^4 x \left (15 c^3+118 c^2 d x+136 c d^2 x^2+48 d^3 x^3\right )\right )}{\sqrt {a+b x}}-\frac {15 (b c-a d)^{3/2} \left (-63 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 {\frac {b (c+d x)}{b c-a d}}}\right )}{192 b^5 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*((Sqrt[d]*(-945*a^4*d^3 + 105*a^3*b*d^2*(17*c - 3*d*x) + a^2*b^2*d*(-839*c^2 + 637*c*d*x + 126*
d^2*x^2) + a*b^3*(15*c^3 - 337*c^2*d*x - 244*c*d^2*x^2 - 72*d^3*x^3) + b^4*x*(15*c^3 + 118*c^2*d*x + 136*c*d^2
*x^2 + 48*d^3*x^3)))/Sqrt[a + b*x] - (15*(b*c - a*d)^(3/2)*(b^2*c^2 + 14*a*b*c*d - 63*a^2*d^2)*ArcSinh[(Sqrt[d
]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(192*b^5*d^(3/2))

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IntegrateAlgebraic [A]  time = 0.74, size = 393, normalized size = 1.28 \begin {gather*} \frac {\sqrt {c+d x} (b c-a d)^2 \left (-\frac {384 a^2 b^4 d (c+d x)^4}{(a+b x)^4}+\frac {2511 a^2 b^3 d^2 (c+d x)^3}{(a+b x)^3}-\frac {4599 a^2 b^2 d^3 (c+d x)^2}{(a+b x)^2}+\frac {3465 a^2 b d^4 (c+d x)}{a+b x}-945 a^2 d^5+\frac {15 b^5 c^2 (c+d x)^3}{(a+b x)^3}+\frac {73 b^4 c^2 d (c+d x)^2}{(a+b x)^2}-\frac {558 a b^4 c d (c+d x)^3}{(a+b x)^3}-\frac {55 b^3 c^2 d^2 (c+d x)}{a+b x}+\frac {1022 a b^3 c d^2 (c+d x)^2}{(a+b x)^2}-\frac {770 a b^2 c d^3 (c+d x)}{a+b x}+210 a b c d^4+15 b^2 c^2 d^3\right )}{192 b^5 d \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^4}-\frac {5 (b c-a d)^2 \left (-63 a^2 d^2+14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{11/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*Sqrt[c + d*x]*(15*b^2*c^2*d^3 + 210*a*b*c*d^4 - 945*a^2*d^5 - (55*b^3*c^2*d^2*(c + d*x))/(a + b
*x) - (770*a*b^2*c*d^3*(c + d*x))/(a + b*x) + (3465*a^2*b*d^4*(c + d*x))/(a + b*x) + (73*b^4*c^2*d*(c + d*x)^2
)/(a + b*x)^2 + (1022*a*b^3*c*d^2*(c + d*x)^2)/(a + b*x)^2 - (4599*a^2*b^2*d^3*(c + d*x)^2)/(a + b*x)^2 + (15*
b^5*c^2*(c + d*x)^3)/(a + b*x)^3 - (558*a*b^4*c*d*(c + d*x)^3)/(a + b*x)^3 + (2511*a^2*b^3*d^2*(c + d*x)^3)/(a
+ b*x)^3 - (384*a^2*b^4*d*(c + d*x)^4)/(a + b*x)^4))/(192*b^5*d*Sqrt[a + b*x]*(-d + (b*(c + d*x))/(a + b*x))^
4) - (5*(b*c - a*d)^2*(b^2*c^2 + 14*a*b*c*d - 63*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*
x])])/(64*b^(11/2)*d^(3/2))

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fricas [A]  time = 2.05, size = 790, normalized size = 2.58 \begin {gather*} \left [-\frac {15 \, {\left (a b^{4} c^{4} + 12 \, a^{2} b^{3} c^{3} d - 90 \, a^{3} b^{2} c^{2} d^{2} + 140 \, a^{4} b c d^{3} - 63 \, a^{5} d^{4} + {\left (b^{5} c^{4} + 12 \, a b^{4} c^{3} d - 90 \, a^{2} b^{3} c^{2} d^{2} + 140 \, a^{3} b^{2} c d^{3} - 63 \, a^{4} b d^{4}\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^{5} d^{4} x^{4} + 15 \, a b^{4} c^{3} d - 839 \, a^{2} b^{3} c^{2} d^{2} + 1785 \, a^{3} b^{2} c d^{3} - 945 \, a^{4} b d^{4} + 8 \, {\left (17 \, b^{5} c d^{3} - 9 \, a b^{4} d^{4}\right )} x^{3} + 2 \, {\left (59 \, b^{5} c^{2} d^{2} - 122 \, a b^{4} c d^{3} + 63 \, a^{2} b^{3} d^{4}\right )} x^{2} + {\left (15 \, b^{5} c^{3} d - 337 \, a b^{4} c^{2} d^{2} + 637 \, a^{2} b^{3} c d^{3} - 315 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, {\left (b^{7} d^{2} x + a b^{6} d^{2}\right )}}, \frac {15 \, {\left (a b^{4} c^{4} + 12 \, a^{2} b^{3} c^{3} d - 90 \, a^{3} b^{2} c^{2} d^{2} + 140 \, a^{4} b c d^{3} - 63 \, a^{5} d^{4} + {\left (b^{5} c^{4} + 12 \, a b^{4} c^{3} d - 90 \, a^{2} b^{3} c^{2} d^{2} + 140 \, a^{3} b^{2} c d^{3} - 63 \, a^{4} b d^{4}\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^{5} d^{4} x^{4} + 15 \, a b^{4} c^{3} d - 839 \, a^{2} b^{3} c^{2} d^{2} + 1785 \, a^{3} b^{2} c d^{3} - 945 \, a^{4} b d^{4} + 8 \, {\left (17 \, b^{5} c d^{3} - 9 \, a b^{4} d^{4}\right )} x^{3} + 2 \, {\left (59 \, b^{5} c^{2} d^{2} - 122 \, a b^{4} c d^{3} + 63 \, a^{2} b^{3} d^{4}\right )} x^{2} + {\left (15 \, b^{5} c^{3} d - 337 \, a b^{4} c^{2} d^{2} + 637 \, a^{2} b^{3} c d^{3} - 315 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, {\left (b^{7} d^{2} x + a b^{6} d^{2}\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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.26, size = 463, normalized size = 1.51 \begin {gather*} \frac {1}{192} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{7}} + \frac {17 \, b^{28} c d^{7} {\left | b \right |} - 33 \, a b^{27} d^{8} {\left | b \right |}}{b^{34} d^{6}}\right )} + \frac {59 \, b^{29} c^{2} d^{6} {\left | b \right |} - 326 \, a b^{28} c d^{7} {\left | b \right |} + 315 \, a^{2} b^{27} d^{8} {\left | b \right |}}{b^{34} d^{6}}\right )} + \frac {3 \, {\left (5 \, b^{30} c^{3} d^{5} {\left | b \right |} - 191 \, a b^{29} c^{2} d^{6} {\left | b \right |} + 511 \, a^{2} b^{28} c d^{7} {\left | b \right |} - 325 \, a^{3} b^{27} d^{8} {\left | b \right |}\right )}}{b^{34} d^{6}}\right )} \sqrt {b x + a} - \frac {4 \, {\left (\sqrt {b d} a^{2} b^{3} c^{3} {\left | b \right |} - 3 \, \sqrt {b d} a^{3} b^{2} c^{2} d {\left | b \right |} + 3 \, \sqrt {b d} a^{4} b c d^{2} {\left | b \right |} - \sqrt {b d} a^{5} d^{3} {\left | b \right |}\right )}}{{\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 )} b^{6}} + \frac {5 \, {\left (\sqrt {b d} b^{4} c^{4} {\left | b \right |} + 12 \, \sqrt {b d} a b^{3} c^{3} d {\left | b \right |} - 90 \, \sqrt {b d} a^{2} b^{2} c^{2} d^{2} {\left | b \right |} + 140 \, \sqrt {b d} a^{3} b c d^{3} {\left | b \right |} - 63 \, \sqrt {b d} a^{4} d^{4} {\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 )}{128 \, b^{7} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)*d^2*abs(b)/b^7 + (17*b^28*c*d
^7*abs(b) - 33*a*b^27*d^8*abs(b))/(b^34*d^6)) + (59*b^29*c^2*d^6*abs(b) - 326*a*b^28*c*d^7*abs(b) + 315*a^2*b^
27*d^8*abs(b))/(b^34*d^6)) + 3*(5*b^30*c^3*d^5*abs(b) - 191*a*b^29*c^2*d^6*abs(b) + 511*a^2*b^28*c*d^7*abs(b)
- 325*a^3*b^27*d^8*abs(b))/(b^34*d^6))*sqrt(b*x + a) - 4*(sqrt(b*d)*a^2*b^3*c^3*abs(b) - 3*sqrt(b*d)*a^3*b^2*c
^2*d*abs(b) + 3*sqrt(b*d)*a^4*b*c*d^2*abs(b) - sqrt(b*d)*a^5*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)*b^6) + 5/128*(sqrt(b*d)*b^4*c^4*abs(b) + 12*sqrt(b*d)*a*b^3*c^
3*d*abs(b) - 90*sqrt(b*d)*a^2*b^2*c^2*d^2*abs(b) + 140*sqrt(b*d)*a^3*b*c*d^3*abs(b) - 63*sqrt(b*d)*a^4*d^4*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^2)

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maple [B]  time = 0.03, size = 961, normalized size = 3.14 \begin {gather*} \frac {\sqrt {d x +c}\, \left (945 a^{4} b \,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 )-2100 a^{3} b^{2} c \,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 )+1350 a^{2} b^{3} c^{2} 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 )-180 a \,b^{4} c^{3} 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 )-15 b^{5} c^{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 )+96 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} d^{3} x^{4}+945 a^{5} 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 )-2100 a^{4} b c \,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^{3} b^{2} c^{2} 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 )-180 a^{2} b^{3} c^{3} 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 \,b^{4} c^{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 )-144 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} d^{3} x^{3}+272 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c \,d^{2} x^{3}+252 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} d^{3} x^{2}-488 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c \,d^{2} x^{2}+236 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{2} d \,x^{2}-630 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} b \,d^{3} x +1274 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c \,d^{2} x -674 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{2} d x +30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{4} c^{3} x -1890 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{4} d^{3}+3570 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} b c \,d^{2}-1678 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b^{2} c^{2} d +30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{3} c^{3}\right )}{384 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{5} 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)^(3/2),x)

[Out]

1/384*(d*x+c)^(1/2)*(96*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^4*d^3*x^4-144*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2
)*a*b^3*d^3*x^3+272*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^4*c*d^2*x^3+945*a^4*b*d^4*x*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))-2100*a^3*b^2*c*d^3*x*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))+1350*a^2*b^3*c^2*d^2*x*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))-180*a*b^4*c^3*d*x*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))-15*b^5*c^4*x*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))+252*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b^2*d^3*x^2-488*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b^3*c*d^2*x^2+236*((b
*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^4*c^2*d*x^2+945*a^5*d^4*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))-2100*a^4*b*c*d^3*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))+1350*a^3*b^2*c^2*d^2*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))-180*a
^2*b^3*c^3*d*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))-15*a*b^4*c^4*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))-630*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^
3*b*d^3*x+1274*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b^2*c*d^2*x-674*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b
^3*c^2*d*x+30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^4*c^3*x-1890*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^4*d^3+3
570*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^3*b*c*d^2-1678*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b^2*c^2*d+30*
((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b^3*c^3)/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^5/d

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (c + d x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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