3.8.60 \(\int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\) [760]

Optimal. Leaf size=246 \[ -\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4 d}-\frac {\left (10 a c+\frac {b c^2}{d}-\frac {35 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {(b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} d^{3/2}} \]

[Out]

-1/8*(-a*d+b*c)*(-35*a^2*d^2+10*a*b*c*d+b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(9/2)/
d^(3/2)-2*a^2*(d*x+c)^(5/2)/b^2/(-a*d+b*c)/(b*x+a)^(1/2)-1/12*(10*a*c+b*c^2/d-35*a^2*d/b)*(d*x+c)^(3/2)*(b*x+a
)^(1/2)/b^2/(-a*d+b*c)+1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/b^2/d-1/8*(-35*a^2*d^2+10*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2
)*(d*x+c)^(1/2)/b^4/d

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Rubi [A]
time = 0.17, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {91, 81, 52, 65, 223, 212} \begin {gather*} -\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-\frac {35 a^2 d}{b}+10 a c+\frac {b c^2}{d}\right )}{12 b^2 (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 \sqrt {a+b x} (b c-a d)}-\frac {(b c-a d) \left (-35 a^2 d^2+10 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^{9/2} d^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-35 a^2 d^2+10 a b c d+b^2 c^2\right )}{8 b^4 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(b^4*d) - ((10*a*c + (b*c^2)/d - (35*a^
2*d)/b)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b^2*(b*c - a*d)) - (2*a^2*(c + d*x)^(5/2))/(b^2*(b*c - a*d)*Sqrt[a
+ b*x]) + (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b^2*d) - ((b*c - a*d)*(b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*ArcTanh
[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(9/2)*d^(3/2))

Rule 52

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 65

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 81

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 91

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 212

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

Rule 223

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)^{3/2}}{(a+b x)^{3/2}} \, dx &=-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {2 \int \frac {(c+d x)^{3/2} \left (-\frac {1}{2} a (b c-5 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)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{6 b^2 d (b c-a d)}\\ &=-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{8 b^3 d}\\ &=-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4 d}-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {\left ((b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b^4 d}\\ &=-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4 d}-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {\left ((b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right )\right ) \text {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^5 d}\\ &=-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4 d}-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {\left ((b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right )\right ) \text {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^5 d}\\ &=-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4 d}-\frac {\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3 d (b c-a d)}-\frac {2 a^2 (c+d x)^{5/2}}{b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 d}-\frac {(b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.26, size = 183, normalized size = 0.74 \begin {gather*} \frac {\frac {b^2 \sqrt {c+d x} \left (105 a^3 d^2+5 a^2 b d (-20 c+7 d x)+a b^2 \left (3 c^2-38 c d x-14 d^2 x^2\right )+b^3 x \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )}{d \sqrt {a+b x}}+3 \left (\frac {b}{d}\right )^{3/2} \left (b^3 c^3+9 a b^2 c^2 d-45 a^2 b c d^2+35 a^3 d^3\right ) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{24 b^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*Sqrt[c + d*x]*(105*a^3*d^2 + 5*a^2*b*d*(-20*c + 7*d*x) + a*b^2*(3*c^2 - 38*c*d*x - 14*d^2*x^2) + b^3*x*(
3*c^2 + 14*c*d*x + 8*d^2*x^2)))/(d*Sqrt[a + b*x]) + 3*(b/d)^(3/2)*(b^3*c^3 + 9*a*b^2*c^2*d - 45*a^2*b*c*d^2 +
35*a^3*d^3)*Log[Sqrt[a + b*x] - Sqrt[b/d]*Sqrt[c + d*x]])/(24*b^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(691\) vs. \(2(210)=420\).
time = 0.08, size = 692, normalized size = 2.81

method	result	size
default	\(-\frac {\sqrt {d x +c}\, \left (-16 b^{3} d^{2} x^{3} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b \,d^{3} x -135 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c \,d^{2} x +27 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{2} d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{4} c^{3} x +28 a \,b^{2} d^{2} x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-28 b^{3} c d \,x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{4} d^{3}-135 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b c \,d^{2}+27 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{3}-70 a^{2} b \,d^{2} x \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+76 a \,b^{2} c d x \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-6 b^{3} c^{2} x \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-210 a^{3} d^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+200 a^{2} b c d \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-6 a \,b^{2} c^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\right )}{48 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{4} d}\)	\(692\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(d*x+c)^(3/2)/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/48*(d*x+c)^(1/2)*(-16*b^3*d^2*x^3*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+105*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a
))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*b*d^3*x-135*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/
2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c*d^2*x+27*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*
d)^(1/2))*a*b^3*c^2*d*x+3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^4*c^3*
x+28*a*b^2*d^2*x^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)-28*b^3*c*d*x^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+105*
ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^4*d^3-135*ln(1/2*(2*b*d*x+2*((d*
x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*b*c*d^2+27*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)
*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c^2*d+3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b
*c)/(b*d)^(1/2))*a*b^3*c^3-70*a^2*b*d^2*x*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+76*a*b^2*c*d*x*((d*x+c)*(b*x+a))
^(1/2)*(b*d)^(1/2)-6*b^3*c^2*x*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)-210*a^3*d^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(
1/2)+200*a^2*b*c*d*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)-6*a*b^2*c^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2))/((d*x+
c)*(b*x+a))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^4/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)^(3/2)/(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 detail

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Fricas [A]
time = 0.97, size = 600, normalized size = 2.44 \begin {gather*} \left [\frac {3 \, {\left (a b^{3} c^{3} + 9 \, a^{2} b^{2} c^{2} d - 45 \, a^{3} b c d^{2} + 35 \, a^{4} d^{3} + {\left (b^{4} c^{3} + 9 \, a b^{3} c^{2} d - 45 \, a^{2} b^{2} c d^{2} + 35 \, a^{3} 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^{4} d^{3} x^{3} + 3 \, a b^{3} c^{2} d - 100 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 14 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + {\left (3 \, b^{4} c^{2} d - 38 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{6} d^{2} x + a b^{5} d^{2}\right )}}, \frac {3 \, {\left (a b^{3} c^{3} + 9 \, a^{2} b^{2} c^{2} d - 45 \, a^{3} b c d^{2} + 35 \, a^{4} d^{3} + {\left (b^{4} c^{3} + 9 \, a b^{3} c^{2} d - 45 \, a^{2} b^{2} c d^{2} + 35 \, a^{3} 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^{4} d^{3} x^{3} + 3 \, a b^{3} c^{2} d - 100 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 14 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + {\left (3 \, b^{4} c^{2} d - 38 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{6} d^{2} x + a b^{5} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(a*b^3*c^3 + 9*a^2*b^2*c^2*d - 45*a^3*b*c*d^2 + 35*a^4*d^3 + (b^4*c^3 + 9*a*b^3*c^2*d - 45*a^2*b^2*c*
d^2 + 35*a^3*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)*s
qrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8*b^4*d^3*x^3 + 3*a*b^3*c^2*d - 100*a^2*b
^2*c*d^2 + 105*a^3*b*d^3 + 14*(b^4*c*d^2 - a*b^3*d^3)*x^2 + (3*b^4*c^2*d - 38*a*b^3*c*d^2 + 35*a^2*b^2*d^3)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d^2*x + a*b^5*d^2), 1/48*(3*(a*b^3*c^3 + 9*a^2*b^2*c^2*d - 45*a^3*b*c*d^2 +
35*a^4*d^3 + (b^4*c^3 + 9*a*b^3*c^2*d - 45*a^2*b^2*c*d^2 + 35*a^3*b*d^3)*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*(8*b^4*
d^3*x^3 + 3*a*b^3*c^2*d - 100*a^2*b^2*c*d^2 + 105*a^3*b*d^3 + 14*(b^4*c*d^2 - a*b^3*d^3)*x^2 + (3*b^4*c^2*d -
38*a*b^3*c*d^2 + 35*a^2*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d^2*x + a*b^5*d^2)]

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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 {3}{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)**(3/2)/(b*x+a)**(3/2),x)

[Out]

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

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Giac [A]
time = 0.96, size = 347, normalized size = 1.41 \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 {\left | b \right |}}{b^{6}} + \frac {7 \, b^{18} c d^{4} {\left | b \right |} - 19 \, a b^{17} d^{5} {\left | b \right |}}{b^{23} d^{4}}\right )} + \frac {3 \, {\left (b^{19} c^{2} d^{3} {\left | b \right |} - 22 \, a b^{18} c d^{4} {\left | b \right |} + 29 \, a^{2} b^{17} d^{5} {\left | b \right |}\right )}}{b^{23} d^{4}}\right )} - \frac {4 \, {\left (\sqrt {b d} a^{2} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a^{3} b c d {\left | b \right |} + \sqrt {b d} a^{4} d^{2} {\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^{5}} + \frac {{\left (\sqrt {b d} b^{3} c^{3} {\left | b \right |} + 9 \, \sqrt {b d} a b^{2} c^{2} d {\left | b \right |} - 45 \, \sqrt {b d} a^{2} b c d^{2} {\left | b \right |} + 35 \, \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^{6} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x+c)^(3/2)/(b*x+a)^(3/2),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*abs(b)/b^6 + (7*b^18*c*d^4*
abs(b) - 19*a*b^17*d^5*abs(b))/(b^23*d^4)) + 3*(b^19*c^2*d^3*abs(b) - 22*a*b^18*c*d^4*abs(b) + 29*a^2*b^17*d^5
*abs(b))/(b^23*d^4)) - 4*(sqrt(b*d)*a^2*b^2*c^2*abs(b) - 2*sqrt(b*d)*a^3*b*c*d*abs(b) + sqrt(b*d)*a^4*d^2*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^5) + 1/16*(sqrt(b*d
)*b^3*c^3*abs(b) + 9*sqrt(b*d)*a*b^2*c^2*d*abs(b) - 45*sqrt(b*d)*a^2*b*c*d^2*abs(b) + 35*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^6*d^2)

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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 )}^{3/2}}{{\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)^(3/2))/(a + b*x)^(3/2),x)

[Out]

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

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