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

Optimal. Leaf size=377 \[ \frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{24 b^4 d (b c-a d)}-\frac {5 (b c-a d) \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{13/2} d^{3/2}}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right )}{64 b^6 d}-\frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right )}{96 b^5 d (b c-a d)}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{3 b^2 \sqrt {a+b x} (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}} \]

[Out]

-2/3*x^3*(d*x+c)^(5/2)/b/(b*x+a)^(3/2)-5/64*(-a*d+b*c)*(231*a^3*d^3-189*a^2*b*c*d^2+21*a*b^2*c^2*d+b^3*c^3)*ar
ctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(13/2)/d^(3/2)-2/3*(-11*a*d+6*b*c)*x^2*(d*x+c)^(5/2)/b^2/
(-a*d+b*c)/(b*x+a)^(1/2)-5/96*(231*a^3*d^3-189*a^2*b*c*d^2+21*a*b^2*c^2*d+b^3*c^3)*(d*x+c)^(3/2)*(b*x+a)^(1/2)
/b^5/d/(-a*d+b*c)+1/24*(d*x+c)^(5/2)*(5*b^2*c^2-156*a*b*c*d+231*a^2*d^2+2*b*d*(-99*a*d+59*b*c)*x)*(b*x+a)^(1/2
)/b^4/d/(-a*d+b*c)-5/64*(231*a^3*d^3-189*a^2*b*c*d^2+21*a*b^2*c^2*d+b^3*c^3)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^6/d

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Rubi [A]  time = 0.36, antiderivative size = 377, 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, 150, 147, 50, 63, 217, 206} \[ -\frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (-189 a^2 b c d^2+231 a^3 d^3+21 a b^2 c^2 d+b^3 c^3\right )}{96 b^5 d (b c-a d)}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (231 a^2 d^2+2 b d x (59 b c-99 a d)-156 a b c d+5 b^2 c^2\right )}{24 b^4 d (b c-a d)}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} \left (-189 a^2 b c d^2+231 a^3 d^3+21 a b^2 c^2 d+b^3 c^3\right )}{64 b^6 d}-\frac {5 (b c-a d) \left (-189 a^2 b c d^2+231 a^3 d^3+21 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{13/2} d^{3/2}}-\frac {2 x^2 (c+d x)^{5/2} (6 b c-11 a d)}{3 b^2 \sqrt {a+b x} (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b^3*c^3 + 21*a*b^2*c^2*d - 189*a^2*b*c*d^2 + 231*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^6*d) - (5*(b
^3*c^3 + 21*a*b^2*c^2*d - 189*a^2*b*c*d^2 + 231*a^3*d^3)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(96*b^5*d*(b*c - a*d))
- (2*x^3*(c + d*x)^(5/2))/(3*b*(a + b*x)^(3/2)) - (2*(6*b*c - 11*a*d)*x^2*(c + d*x)^(5/2))/(3*b^2*(b*c - a*d)
*Sqrt[a + b*x]) + (Sqrt[a + b*x]*(c + d*x)^(5/2)*(5*b^2*c^2 - 156*a*b*c*d + 231*a^2*d^2 + 2*b*d*(59*b*c - 99*a
*d)*x))/(24*b^4*d*(b*c - a*d)) - (5*(b*c - a*d)*(b^3*c^3 + 21*a*b^2*c^2*d - 189*a^2*b*c*d^2 + 231*a^3*d^3)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(13/2)*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 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 150

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

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)^{5/2}} \, dx &=-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {2 \int \frac {x^2 (c+d x)^{3/2} \left (3 c+\frac {11 d x}{2}\right )}{(a+b x)^{3/2}} \, dx}{3 b}\\ &=-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {4 \int \frac {x (c+d x)^{3/2} \left (c (6 b c-11 a d)+\frac {1}{4} d (59 b c-99 a d) x\right )}{\sqrt {a+b x}} \, dx}{3 b^2 (b c-a d)}\\ &=-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2-156 a b c d+231 a^2 d^2+2 b d (59 b c-99 a d) x\right )}{24 b^4 d (b c-a d)}-\frac {\left (5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right )\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{48 b^4 d (b c-a d)}\\ &=-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^5 d (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2-156 a b c d+231 a^2 d^2+2 b d (59 b c-99 a d) x\right )}{24 b^4 d (b c-a d)}-\frac {\left (5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{64 b^5 d}\\ &=-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^6 d}-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^5 d (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2-156 a b c d+231 a^2 d^2+2 b d (59 b c-99 a d) x\right )}{24 b^4 d (b c-a d)}-\frac {\left (5 (b c-a d) \left (b^3 c^3+21 a b^2 c^2 d-189 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}{128 b^6 d}\\ &=-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^6 d}-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^5 d (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2-156 a b c d+231 a^2 d^2+2 b d (59 b c-99 a d) x\right )}{24 b^4 d (b c-a d)}-\frac {\left (5 (b c-a d) \left (b^3 c^3+21 a b^2 c^2 d-189 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 )}{64 b^7 d}\\ &=-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^6 d}-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^5 d (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2-156 a b c d+231 a^2 d^2+2 b d (59 b c-99 a d) x\right )}{24 b^4 d (b c-a d)}-\frac {\left (5 (b c-a d) \left (b^3 c^3+21 a b^2 c^2 d-189 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 )}{64 b^7 d}\\ &=-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^6 d}-\frac {5 \left (b^3 c^3+21 a b^2 c^2 d-189 a^2 b c d^2+231 a^3 d^3\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^5 d (b c-a d)}-\frac {2 x^3 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}-\frac {2 (6 b c-11 a d) x^2 (c+d x)^{5/2}}{3 b^2 (b c-a d) \sqrt {a+b x}}+\frac {\sqrt {a+b x} (c+d x)^{5/2} \left (5 b^2 c^2-156 a b c d+231 a^2 d^2+2 b d (59 b c-99 a d) x\right )}{24 b^4 d (b c-a d)}-\frac {5 (b c-a d) \left (b^3 c^3+21 a b^2 c^2 d-189 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 )}{64 b^{13/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 302, normalized size = 0.80 \[ \frac {\sqrt {c+d x} \left (\frac {\sqrt {d} \left (-3465 a^5 d^3+105 a^4 b d^2 (49 c-44 d x)-21 a^3 b^2 d \left (83 c^2-334 c d x+33 d^2 x^2\right )+3 a^2 b^3 \left (5 c^3-824 c^2 d x+387 c d^2 x^2+66 d^3 x^3\right )-a b^4 x \left (-30 c^3+483 c^2 d x+316 c d^2 x^2+88 d^3 x^3\right )+b^5 x^2 \left (15 c^3+118 c^2 d x+136 c d^2 x^2+48 d^3 x^3\right )\right )}{(a+b x)^{3/2}}-\frac {15 \sqrt {b c-a d} \left (231 a^3 d^3-189 a^2 b c d^2+21 a b^2 c^2 d+b^3 c^3\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{192 b^6 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*((Sqrt[d]*(-3465*a^5*d^3 + 105*a^4*b*d^2*(49*c - 44*d*x) - 21*a^3*b^2*d*(83*c^2 - 334*c*d*x + 3
3*d^2*x^2) + b^5*x^2*(15*c^3 + 118*c^2*d*x + 136*c*d^2*x^2 + 48*d^3*x^3) + 3*a^2*b^3*(5*c^3 - 824*c^2*d*x + 38
7*c*d^2*x^2 + 66*d^3*x^3) - a*b^4*x*(-30*c^3 + 483*c^2*d*x + 316*c*d^2*x^2 + 88*d^3*x^3)))/(a + b*x)^(3/2) - (
15*Sqrt[b*c - a*d]*(b^3*c^3 + 21*a*b^2*c^2*d - 189*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)]))/(192*b^6*d^(3/2))

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fricas [A]  time = 2.15, size = 1060, normalized size = 2.81 \[ \left [-\frac {15 \, {\left (a^{2} b^{4} c^{4} + 20 \, a^{3} b^{3} c^{3} d - 210 \, a^{4} b^{2} c^{2} d^{2} + 420 \, a^{5} b c d^{3} - 231 \, a^{6} d^{4} + {\left (b^{6} c^{4} + 20 \, a b^{5} c^{3} d - 210 \, a^{2} b^{4} c^{2} d^{2} + 420 \, a^{3} b^{3} c d^{3} - 231 \, a^{4} b^{2} d^{4}\right )} x^{2} + 2 \, {\left (a b^{5} c^{4} + 20 \, a^{2} b^{4} c^{3} d - 210 \, a^{3} b^{3} c^{2} d^{2} + 420 \, a^{4} b^{2} c d^{3} - 231 \, a^{5} 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^{6} d^{4} x^{5} + 15 \, a^{2} b^{4} c^{3} d - 1743 \, a^{3} b^{3} c^{2} d^{2} + 5145 \, a^{4} b^{2} c d^{3} - 3465 \, a^{5} b d^{4} + 8 \, {\left (17 \, b^{6} c d^{3} - 11 \, a b^{5} d^{4}\right )} x^{4} + 2 \, {\left (59 \, b^{6} c^{2} d^{2} - 158 \, a b^{5} c d^{3} + 99 \, a^{2} b^{4} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{6} c^{3} d - 161 \, a b^{5} c^{2} d^{2} + 387 \, a^{2} b^{4} c d^{3} - 231 \, a^{3} b^{3} d^{4}\right )} x^{2} + 6 \, {\left (5 \, a b^{5} c^{3} d - 412 \, a^{2} b^{4} c^{2} d^{2} + 1169 \, a^{3} b^{3} c d^{3} - 770 \, a^{4} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, {\left (b^{9} d^{2} x^{2} + 2 \, a b^{8} d^{2} x + a^{2} b^{7} d^{2}\right )}}, \frac {15 \, {\left (a^{2} b^{4} c^{4} + 20 \, a^{3} b^{3} c^{3} d - 210 \, a^{4} b^{2} c^{2} d^{2} + 420 \, a^{5} b c d^{3} - 231 \, a^{6} d^{4} + {\left (b^{6} c^{4} + 20 \, a b^{5} c^{3} d - 210 \, a^{2} b^{4} c^{2} d^{2} + 420 \, a^{3} b^{3} c d^{3} - 231 \, a^{4} b^{2} d^{4}\right )} x^{2} + 2 \, {\left (a b^{5} c^{4} + 20 \, a^{2} b^{4} c^{3} d - 210 \, a^{3} b^{3} c^{2} d^{2} + 420 \, a^{4} b^{2} c d^{3} - 231 \, a^{5} 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^{6} d^{4} x^{5} + 15 \, a^{2} b^{4} c^{3} d - 1743 \, a^{3} b^{3} c^{2} d^{2} + 5145 \, a^{4} b^{2} c d^{3} - 3465 \, a^{5} b d^{4} + 8 \, {\left (17 \, b^{6} c d^{3} - 11 \, a b^{5} d^{4}\right )} x^{4} + 2 \, {\left (59 \, b^{6} c^{2} d^{2} - 158 \, a b^{5} c d^{3} + 99 \, a^{2} b^{4} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{6} c^{3} d - 161 \, a b^{5} c^{2} d^{2} + 387 \, a^{2} b^{4} c d^{3} - 231 \, a^{3} b^{3} d^{4}\right )} x^{2} + 6 \, {\left (5 \, a b^{5} c^{3} d - 412 \, a^{2} b^{4} c^{2} d^{2} + 1169 \, a^{3} b^{3} c d^{3} - 770 \, a^{4} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, {\left (b^{9} d^{2} x^{2} + 2 \, a b^{8} d^{2} x + a^{2} b^{7} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(15*(a^2*b^4*c^4 + 20*a^3*b^3*c^3*d - 210*a^4*b^2*c^2*d^2 + 420*a^5*b*c*d^3 - 231*a^6*d^4 + (b^6*c^4 +
20*a*b^5*c^3*d - 210*a^2*b^4*c^2*d^2 + 420*a^3*b^3*c*d^3 - 231*a^4*b^2*d^4)*x^2 + 2*(a*b^5*c^4 + 20*a^2*b^4*c
^3*d - 210*a^3*b^3*c^2*d^2 + 420*a^4*b^2*c*d^3 - 231*a^5*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^6*d^4*x^5 + 15*a^2*b^4*c^3*d - 1743*a^3*b^3*c^2*d^2 + 5145*a^4*b^2*c*d^3 - 3465*a^5*b*d^4 + 8*(17*b^6*
c*d^3 - 11*a*b^5*d^4)*x^4 + 2*(59*b^6*c^2*d^2 - 158*a*b^5*c*d^3 + 99*a^2*b^4*d^4)*x^3 + 3*(5*b^6*c^3*d - 161*a
*b^5*c^2*d^2 + 387*a^2*b^4*c*d^3 - 231*a^3*b^3*d^4)*x^2 + 6*(5*a*b^5*c^3*d - 412*a^2*b^4*c^2*d^2 + 1169*a^3*b^
3*c*d^3 - 770*a^4*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^9*d^2*x^2 + 2*a*b^8*d^2*x + a^2*b^7*d^2), 1/384*
(15*(a^2*b^4*c^4 + 20*a^3*b^3*c^3*d - 210*a^4*b^2*c^2*d^2 + 420*a^5*b*c*d^3 - 231*a^6*d^4 + (b^6*c^4 + 20*a*b^
5*c^3*d - 210*a^2*b^4*c^2*d^2 + 420*a^3*b^3*c*d^3 - 231*a^4*b^2*d^4)*x^2 + 2*(a*b^5*c^4 + 20*a^2*b^4*c^3*d - 2
10*a^3*b^3*c^2*d^2 + 420*a^4*b^2*c*d^3 - 231*a^5*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^6*d^4*x^5 + 15*a^2*
b^4*c^3*d - 1743*a^3*b^3*c^2*d^2 + 5145*a^4*b^2*c*d^3 - 3465*a^5*b*d^4 + 8*(17*b^6*c*d^3 - 11*a*b^5*d^4)*x^4 +
2*(59*b^6*c^2*d^2 - 158*a*b^5*c*d^3 + 99*a^2*b^4*d^4)*x^3 + 3*(5*b^6*c^3*d - 161*a*b^5*c^2*d^2 + 387*a^2*b^4*
c*d^3 - 231*a^3*b^3*d^4)*x^2 + 6*(5*a*b^5*c^3*d - 412*a^2*b^4*c^2*d^2 + 1169*a^3*b^3*c*d^3 - 770*a^4*b^2*d^4)*
x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^9*d^2*x^2 + 2*a*b^8*d^2*x + a^2*b^7*d^2)]

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giac [B]  time = 3.83, size = 1024, normalized size = 2.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(d*x+c)^(5/2)/(b*x+a)^(5/2),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^8 + (17*b^32*c*d
^7*abs(b) - 41*a*b^31*d^8*abs(b))/(b^39*d^6)) + (59*b^33*c^2*d^6*abs(b) - 430*a*b^32*c*d^7*abs(b) + 515*a^2*b^
31*d^8*abs(b))/(b^39*d^6)) + 3*(5*b^34*c^3*d^5*abs(b) - 279*a*b^33*c^2*d^6*abs(b) + 975*a^2*b^32*c*d^7*abs(b)
- 765*a^3*b^31*d^8*abs(b))/(b^39*d^6))*sqrt(b*x + a) + 5/128*(sqrt(b*d)*b^4*c^4*abs(b) + 20*sqrt(b*d)*a*b^3*c^
3*d*abs(b) - 210*sqrt(b*d)*a^2*b^2*c^2*d^2*abs(b) + 420*sqrt(b*d)*a^3*b*c*d^3*abs(b) - 231*sqrt(b*d)*a^4*d^4*a
bs(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^2) - 4/3*(9*sqrt(b*d)*a^2
*b^7*c^5*abs(b) - 52*sqrt(b*d)*a^3*b^6*c^4*d*abs(b) + 118*sqrt(b*d)*a^4*b^5*c^3*d^2*abs(b) - 132*sqrt(b*d)*a^5
*b^4*c^2*d^3*abs(b) + 73*sqrt(b*d)*a^6*b^3*c*d^4*abs(b) - 16*sqrt(b*d)*a^7*b^2*d^5*abs(b) - 18*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^5*c^4*abs(b) + 84*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^4*c^3*d*abs(b) - 144*sqrt(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^3*c^2*d^2*abs(b) + 108*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^2*c*d^3*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^2*a^6*b*d^4*abs(b) + 9*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^3*c^3*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^2*c^2*d*abs(b) + 45*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*b*c*d^2*abs(b) - 18*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*d^3*ab
s(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^7)

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maple [B]  time = 0.03, size = 1366, normalized size = 3.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(d*x+c)^(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))*a^6*d^4-
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^4*c^4-15*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^6*c^4-6930*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/
2)*a^5*d^3-600*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+346
5*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^4*b^2*d^4-300*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^5*c^3*d+60*((b*x+a)*(d*x+c))^(1/2)*(b*
d)^(1/2)*x*a*b^4*c^3-3486*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^3*b^2*c^2*d+396*((b*x+a)*(d*x+c))^(1/2)*(b*d)^
(1/2)*x^3*a^2*b^3*d^3+236*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^3*b^5*c^2*d-176*x^4*a*b^4*d^3*(b*d)^(1/2)*((b*
x+a)*(d*x+c))^(1/2)+272*x^4*b^5*c*d^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-1386*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(
1/2)*x^2*a^3*b^2*d^3-12600*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^3+6300*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^2-92
40*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^4*b*d^3+10290*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^4*b*c*d^2-6300*
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^3*c*d^3+3150*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^4*c^2*d^2+6930*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^4-6300*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^3+3150*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^2-30*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^5*c^4-300*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+30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^2*b^5*c^3+30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b^3*c^3+96*x^
5*b^5*d^3*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2322*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^2*b^3*c*d^2+14028
*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a^3*b^2*c*d^2-632*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^3*a*b^4*c*d^2-9
66*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x^2*a*b^4*c^2*d-4944*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*x*a^2*b^3*c^2*
d)/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(3/2)/b^6/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^3*(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 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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