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

Optimal. Leaf size=346 \[ -\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (-231 a^2 d^2-2 b d x (5 b c-99 a d)+30 a b c d+5 b^2 c^2\right )}{80 b^4 d^2}+\frac {3 (b c-a d)^2 \left (-231 a^3 d^3+63 a^2 b c d^2+7 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{13/2} d^{5/2}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) \left (-231 a^3 d^3+63 a^2 b c d^2+7 a b^2 c^2 d+b^3 c^3\right )}{128 b^6 d^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-231 a^3 d^3+63 a^2 b c d^2+7 a b^2 c^2 d+b^3 c^3\right )}{64 b^5 d^2}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}} \]

[Out]

3/128*(-a*d+b*c)^2*(-231*a^3*d^3+63*a^2*b*c*d^2+7*a*b^2*c^2*d+b^3*c^3)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(
d*x+c)^(1/2))/b^(13/2)/d^(5/2)-2*x^3*(d*x+c)^(5/2)/b/(b*x+a)^(1/2)+1/64*(-231*a^3*d^3+63*a^2*b*c*d^2+7*a*b^2*c
^2*d+b^3*c^3)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b^5/d^2+11/5*x^2*(d*x+c)^(5/2)*(b*x+a)^(1/2)/b^2-1/80*(d*x+c)^(5/2)*
(5*b^2*c^2+30*a*b*c*d-231*a^2*d^2-2*b*d*(-99*a*d+5*b*c)*x)*(b*x+a)^(1/2)/b^4/d^2+3/128*(-a*d+b*c)*(-231*a^3*d^
3+63*a^2*b*c*d^2+7*a*b^2*c^2*d+b^3*c^3)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^6/d^2

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Rubi [A]  time = 0.30, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {97, 153, 147, 50, 63, 217, 206} \[ \frac {\sqrt {a+b x} (c+d x)^{3/2} \left (63 a^2 b c d^2-231 a^3 d^3+7 a b^2 c^2 d+b^3 c^3\right )}{64 b^5 d^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (-231 a^2 d^2-2 b d x (5 b c-99 a d)+30 a b c d+5 b^2 c^2\right )}{80 b^4 d^2}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) \left (63 a^2 b c d^2-231 a^3 d^3+7 a b^2 c^2 d+b^3 c^3\right )}{128 b^6 d^2}+\frac {3 (b c-a d)^2 \left (63 a^2 b c d^2-231 a^3 d^3+7 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{13/2} d^{5/2}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)*(b^3*c^3 + 7*a*b^2*c^2*d + 63*a^2*b*c*d^2 - 231*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/(128*b^6*
d^2) + ((b^3*c^3 + 7*a*b^2*c^2*d + 63*a^2*b*c*d^2 - 231*a^3*d^3)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(64*b^5*d^2) -
 (2*x^3*(c + d*x)^(5/2))/(b*Sqrt[a + b*x]) + (11*x^2*Sqrt[a + b*x]*(c + d*x)^(5/2))/(5*b^2) - (Sqrt[a + b*x]*(
c + d*x)^(5/2)*(5*b^2*c^2 + 30*a*b*c*d - 231*a^2*d^2 - 2*b*d*(5*b*c - 99*a*d)*x))/(80*b^4*d^2) + (3*(b*c - a*d
)^2*(b^3*c^3 + 7*a*b^2*c^2*d + 63*a^2*b*c*d^2 - 231*a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
 d*x])])/(128*b^(13/2)*d^(5/2))

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 97

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
 + p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

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^3 (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx &=-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {2 \int \frac {x^2 (c+d x)^{3/2} \left (3 c+\frac {11 d x}{2}\right )}{\sqrt {a+b x}} \, dx}{b}\\ &=-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}+\frac {2 \int \frac {x (c+d x)^{3/2} \left (-11 a c d+\frac {1}{4} d (5 b c-99 a d) x\right )}{\sqrt {a+b x}} \, dx}{5 b^2 d}\\ &=-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2+30 a b c d-231 a^2 d^2-2 b d (5 b c-99 a d) x\right )}{80 b^4 d^2}+\frac {\left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{32 b^4 d^2}\\ &=\frac {\left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{64 b^5 d^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2+30 a b c d-231 a^2 d^2-2 b d (5 b c-99 a d) x\right )}{80 b^4 d^2}+\frac {\left (3 (b c-a d) \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{128 b^5 d^2}\\ &=\frac {3 (b c-a d) \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{128 b^6 d^2}+\frac {\left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{64 b^5 d^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2+30 a b c d-231 a^2 d^2-2 b d (5 b c-99 a d) x\right )}{80 b^4 d^2}+\frac {\left (3 (b c-a d)^2 \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{256 b^6 d^2}\\ &=\frac {3 (b c-a d) \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{128 b^6 d^2}+\frac {\left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{64 b^5 d^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2+30 a b c d-231 a^2 d^2-2 b d (5 b c-99 a d) x\right )}{80 b^4 d^2}+\frac {\left (3 (b c-a d)^2 \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\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 )}{128 b^7 d^2}\\ &=\frac {3 (b c-a d) \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{128 b^6 d^2}+\frac {\left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{64 b^5 d^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2+30 a b c d-231 a^2 d^2-2 b d (5 b c-99 a d) x\right )}{80 b^4 d^2}+\frac {\left (3 (b c-a d)^2 \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\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 )}{128 b^7 d^2}\\ &=\frac {3 (b c-a d) \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{128 b^6 d^2}+\frac {\left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{64 b^5 d^2}-\frac {2 x^3 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {11 x^2 \sqrt {a+b x} (c+d x)^{5/2}}{5 b^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2+30 a b c d-231 a^2 d^2-2 b d (5 b c-99 a d) x\right )}{80 b^4 d^2}+\frac {3 (b c-a d)^2 \left (b^3 c^3+7 a b^2 c^2 d+63 a^2 b c d^2-231 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{13/2} d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 1.45, size = 324, normalized size = 0.94 \[ \frac {\sqrt {c+d x} \left (\frac {15 \left (-231 a^3 d^3+63 a^2 b c d^2+7 a b^2 c^2 d+b^3 c^3\right ) (b c-a d)^{3/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}+\frac {\sqrt {d} \left (3465 a^5 d^4+105 a^4 b d^3 (11 d x-64 c)-42 a^3 b^2 d^2 \left (-79 c^2+57 c d x+11 d^2 x^2\right )+2 a^2 b^3 d \left (-40 c^3+662 c^2 d x+459 c d^2 x^2+132 d^3 x^3\right )-a b^4 \left (15 c^4+70 c^3 d x+466 c^2 d^2 x^2+512 c d^3 x^3+176 d^4 x^4\right )+b^5 x \left (-15 c^4+10 c^3 d x+248 c^2 d^2 x^2+336 c d^3 x^3+128 d^4 x^4\right )\right )}{\sqrt {a+b x}}\right )}{640 b^6 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*((Sqrt[d]*(3465*a^5*d^4 + 105*a^4*b*d^3*(-64*c + 11*d*x) - 42*a^3*b^2*d^2*(-79*c^2 + 57*c*d*x +
 11*d^2*x^2) + 2*a^2*b^3*d*(-40*c^3 + 662*c^2*d*x + 459*c*d^2*x^2 + 132*d^3*x^3) + b^5*x*(-15*c^4 + 10*c^3*d*x
 + 248*c^2*d^2*x^2 + 336*c*d^3*x^3 + 128*d^4*x^4) - a*b^4*(15*c^4 + 70*c^3*d*x + 466*c^2*d^2*x^2 + 512*c*d^3*x
^3 + 176*d^4*x^4)))/Sqrt[a + b*x] + (15*(b*c - a*d)^(3/2)*(b^3*c^3 + 7*a*b^2*c^2*d + 63*a^2*b*c*d^2 - 231*a^3*
d^3)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(640*b^6*d^(5/2))

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fricas [A]  time = 1.90, size = 1008, normalized size = 2.91 \[ \left [-\frac {15 \, {\left (a b^{5} c^{5} + 5 \, a^{2} b^{4} c^{4} d + 50 \, a^{3} b^{3} c^{3} d^{2} - 350 \, a^{4} b^{2} c^{2} d^{3} + 525 \, a^{5} b c d^{4} - 231 \, a^{6} d^{5} + {\left (b^{6} c^{5} + 5 \, a b^{5} c^{4} d + 50 \, a^{2} b^{4} c^{3} d^{2} - 350 \, a^{3} b^{3} c^{2} d^{3} + 525 \, a^{4} b^{2} c d^{4} - 231 \, a^{5} b 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 (128 \, b^{6} d^{5} x^{5} - 15 \, a b^{5} c^{4} d - 80 \, a^{2} b^{4} c^{3} d^{2} + 3318 \, a^{3} b^{3} c^{2} d^{3} - 6720 \, a^{4} b^{2} c d^{4} + 3465 \, a^{5} b d^{5} + 16 \, {\left (21 \, b^{6} c d^{4} - 11 \, a b^{5} d^{5}\right )} x^{4} + 8 \, {\left (31 \, b^{6} c^{2} d^{3} - 64 \, a b^{5} c d^{4} + 33 \, a^{2} b^{4} d^{5}\right )} x^{3} + 2 \, {\left (5 \, b^{6} c^{3} d^{2} - 233 \, a b^{5} c^{2} d^{3} + 459 \, a^{2} b^{4} c d^{4} - 231 \, a^{3} b^{3} d^{5}\right )} x^{2} - {\left (15 \, b^{6} c^{4} d + 70 \, a b^{5} c^{3} d^{2} - 1324 \, a^{2} b^{4} c^{2} d^{3} + 2394 \, a^{3} b^{3} c d^{4} - 1155 \, a^{4} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2560 \, {\left (b^{8} d^{3} x + a b^{7} d^{3}\right )}}, -\frac {15 \, {\left (a b^{5} c^{5} + 5 \, a^{2} b^{4} c^{4} d + 50 \, a^{3} b^{3} c^{3} d^{2} - 350 \, a^{4} b^{2} c^{2} d^{3} + 525 \, a^{5} b c d^{4} - 231 \, a^{6} d^{5} + {\left (b^{6} c^{5} + 5 \, a b^{5} c^{4} d + 50 \, a^{2} b^{4} c^{3} d^{2} - 350 \, a^{3} b^{3} c^{2} d^{3} + 525 \, a^{4} b^{2} c d^{4} - 231 \, a^{5} b 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 (128 \, b^{6} d^{5} x^{5} - 15 \, a b^{5} c^{4} d - 80 \, a^{2} b^{4} c^{3} d^{2} + 3318 \, a^{3} b^{3} c^{2} d^{3} - 6720 \, a^{4} b^{2} c d^{4} + 3465 \, a^{5} b d^{5} + 16 \, {\left (21 \, b^{6} c d^{4} - 11 \, a b^{5} d^{5}\right )} x^{4} + 8 \, {\left (31 \, b^{6} c^{2} d^{3} - 64 \, a b^{5} c d^{4} + 33 \, a^{2} b^{4} d^{5}\right )} x^{3} + 2 \, {\left (5 \, b^{6} c^{3} d^{2} - 233 \, a b^{5} c^{2} d^{3} + 459 \, a^{2} b^{4} c d^{4} - 231 \, a^{3} b^{3} d^{5}\right )} x^{2} - {\left (15 \, b^{6} c^{4} d + 70 \, a b^{5} c^{3} d^{2} - 1324 \, a^{2} b^{4} c^{2} d^{3} + 2394 \, a^{3} b^{3} c d^{4} - 1155 \, a^{4} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{1280 \, {\left (b^{8} d^{3} x + a b^{7} d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2560*(15*(a*b^5*c^5 + 5*a^2*b^4*c^4*d + 50*a^3*b^3*c^3*d^2 - 350*a^4*b^2*c^2*d^3 + 525*a^5*b*c*d^4 - 231*a
^6*d^5 + (b^6*c^5 + 5*a*b^5*c^4*d + 50*a^2*b^4*c^3*d^2 - 350*a^3*b^3*c^2*d^3 + 525*a^4*b^2*c*d^4 - 231*a^5*b*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*(128*b^6*d^5*x^5 - 15*a*b^5*c^4*d - 80*a^2*b^4*c^3*d^2 + 3
318*a^3*b^3*c^2*d^3 - 6720*a^4*b^2*c*d^4 + 3465*a^5*b*d^5 + 16*(21*b^6*c*d^4 - 11*a*b^5*d^5)*x^4 + 8*(31*b^6*c
^2*d^3 - 64*a*b^5*c*d^4 + 33*a^2*b^4*d^5)*x^3 + 2*(5*b^6*c^3*d^2 - 233*a*b^5*c^2*d^3 + 459*a^2*b^4*c*d^4 - 231
*a^3*b^3*d^5)*x^2 - (15*b^6*c^4*d + 70*a*b^5*c^3*d^2 - 1324*a^2*b^4*c^2*d^3 + 2394*a^3*b^3*c*d^4 - 1155*a^4*b^
2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^8*d^3*x + a*b^7*d^3), -1/1280*(15*(a*b^5*c^5 + 5*a^2*b^4*c^4*d + 50*
a^3*b^3*c^3*d^2 - 350*a^4*b^2*c^2*d^3 + 525*a^5*b*c*d^4 - 231*a^6*d^5 + (b^6*c^5 + 5*a*b^5*c^4*d + 50*a^2*b^4*
c^3*d^2 - 350*a^3*b^3*c^2*d^3 + 525*a^4*b^2*c*d^4 - 231*a^5*b*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*(128*b^6*d^5*x
^5 - 15*a*b^5*c^4*d - 80*a^2*b^4*c^3*d^2 + 3318*a^3*b^3*c^2*d^3 - 6720*a^4*b^2*c*d^4 + 3465*a^5*b*d^5 + 16*(21
*b^6*c*d^4 - 11*a*b^5*d^5)*x^4 + 8*(31*b^6*c^2*d^3 - 64*a*b^5*c*d^4 + 33*a^2*b^4*d^5)*x^3 + 2*(5*b^6*c^3*d^2 -
 233*a*b^5*c^2*d^3 + 459*a^2*b^4*c*d^4 - 231*a^3*b^3*d^5)*x^2 - (15*b^6*c^4*d + 70*a*b^5*c^3*d^2 - 1324*a^2*b^
4*c^2*d^3 + 2394*a^3*b^3*c*d^4 - 1155*a^4*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^8*d^3*x + a*b^7*d^3)]

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giac [A]  time = 3.14, size = 570, normalized size = 1.65 \[ \frac {1}{640} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (4 \, {\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {8 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{8}} + \frac {3 \, {\left (7 \, b^{40} c d^{9} {\left | b \right |} - 17 \, a b^{39} d^{10} {\left | b \right |}\right )}}{b^{47} d^{8}}\right )} + \frac {31 \, b^{41} c^{2} d^{8} {\left | b \right |} - 232 \, a b^{40} c d^{9} {\left | b \right |} + 281 \, a^{2} b^{39} d^{10} {\left | b \right |}}{b^{47} d^{8}}\right )} + \frac {5 \, {\left (b^{42} c^{3} d^{7} {\left | b \right |} - 121 \, a b^{41} c^{2} d^{8} {\left | b \right |} + 447 \, a^{2} b^{40} c d^{9} {\left | b \right |} - 359 \, a^{3} b^{39} d^{10} {\left | b \right |}\right )}}{b^{47} d^{8}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (b^{43} c^{4} d^{6} {\left | b \right |} + 6 \, a b^{42} c^{3} d^{7} {\left | b \right |} - 200 \, a^{2} b^{41} c^{2} d^{8} {\left | b \right |} + 474 \, a^{3} b^{40} c d^{9} {\left | b \right |} - 281 \, a^{4} b^{39} d^{10} {\left | b \right |}\right )}}{b^{47} d^{8}}\right )} \sqrt {b x + a} + \frac {4 \, {\left (\sqrt {b d} a^{3} b^{3} c^{3} {\left | b \right |} - 3 \, \sqrt {b d} a^{4} b^{2} c^{2} d {\left | b \right |} + 3 \, \sqrt {b d} a^{5} b c d^{2} {\left | b \right |} - \sqrt {b d} a^{6} 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^{7}} - \frac {3 \, {\left (\sqrt {b d} b^{5} c^{5} {\left | b \right |} + 5 \, \sqrt {b d} a b^{4} c^{4} d {\left | b \right |} + 50 \, \sqrt {b d} a^{2} b^{3} c^{3} d^{2} {\left | b \right |} - 350 \, \sqrt {b d} a^{3} b^{2} c^{2} d^{3} {\left | b \right |} + 525 \, \sqrt {b d} a^{4} b c d^{4} {\left | b \right |} - 231 \, \sqrt {b d} a^{5} d^{5} {\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 )}{256 \, b^{8} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(b*x + a)*(2*(b*x + a)*(8*(b*x + a)*d^2*abs(b)/b^8 + 3*(7*b^40
*c*d^9*abs(b) - 17*a*b^39*d^10*abs(b))/(b^47*d^8)) + (31*b^41*c^2*d^8*abs(b) - 232*a*b^40*c*d^9*abs(b) + 281*a
^2*b^39*d^10*abs(b))/(b^47*d^8)) + 5*(b^42*c^3*d^7*abs(b) - 121*a*b^41*c^2*d^8*abs(b) + 447*a^2*b^40*c*d^9*abs
(b) - 359*a^3*b^39*d^10*abs(b))/(b^47*d^8))*(b*x + a) - 15*(b^43*c^4*d^6*abs(b) + 6*a*b^42*c^3*d^7*abs(b) - 20
0*a^2*b^41*c^2*d^8*abs(b) + 474*a^3*b^40*c*d^9*abs(b) - 281*a^4*b^39*d^10*abs(b))/(b^47*d^8))*sqrt(b*x + a) +
4*(sqrt(b*d)*a^3*b^3*c^3*abs(b) - 3*sqrt(b*d)*a^4*b^2*c^2*d*abs(b) + 3*sqrt(b*d)*a^5*b*c*d^2*abs(b) - sqrt(b*d
)*a^6*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^7) -
3/256*(sqrt(b*d)*b^5*c^5*abs(b) + 5*sqrt(b*d)*a*b^4*c^4*d*abs(b) + 50*sqrt(b*d)*a^2*b^3*c^3*d^2*abs(b) - 350*s
qrt(b*d)*a^3*b^2*c^2*d^3*abs(b) + 525*sqrt(b*d)*a^4*b*c*d^4*abs(b) - 231*sqrt(b*d)*a^5*d^5*abs(b))*log((sqrt(b
*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^8*d^3)

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maple [B]  time = 0.03, size = 1265, normalized size = 3.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(d*x+c)^(1/2)*(-6930*a^5*d^4*((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*b^6*c^5-15*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*b^5*c^5-256*x^5*b^5*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-6636*a^3*b^2*c^2*d^2*((b*x+
a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*x*b^5*c^4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+30*a*b^4*c^4*((b*x+a)*(d*x+c))^
(1/2)*(b*d)^(1/2)+3465*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^5*b*d^5
-7875*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^5*b*c*d^4+5250*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^2*c^2*d^3-750*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^3*c^3*d^2-75*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^4*c^4*d+3465*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^6*d^5-672*x^4*b^5*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-528*x^3*a^2*b^3*d^4*((b*x+a
)*(d*x+c))^(1/2)*(b*d)^(1/2)-496*x^3*b^5*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+924*x^2*a^3*b^2*d^4*((b*x
+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-20*x^2*b^5*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-2310*x*a^4*b*d^4*((b*x+a)*
(d*x+c))^(1/2)*(b*d)^(1/2)+13440*a^4*b*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-75*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^5*c^4*d+352*x^4*a*b^4*d^4*((b*x+a)*(d*x+c))^(1/2)*(b*d
)^(1/2)-750*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^4*c^3*d^2+5250
*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^3*c^2*d^3+160*a^2*b^3*c^3
*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-7875*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^2*c*d^4+1024*x^3*a*b^4*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-1836*x^2*a^2*b^3*c*d^3*((b*
x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+932*x^2*a*b^4*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+4788*x*a^3*b^2*c*d^3
*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-2648*x*a^2*b^3*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+140*x*a*b^4*c^
3*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^6/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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