\(\int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx\) [1443]

Optimal result

Integrand size = 17, antiderivative size = 200 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=-\frac {35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac {21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {105 b d^3}{8 (b c-a d)^5 \sqrt {c+d x}}+\frac {105 b^{3/2} d^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{11/2}} \]

[Out]

-35/8*d^3/(-a*d+b*c)^4/(d*x+c)^(3/2)-1/3/(-a*d+b*c)/(b*x+a)^3/(d*x+c)^(3/2)+3/4*d/(-a*d+b*c)^2/(b*x+a)^2/(d*x+
c)^(3/2)-21/8*d^2/(-a*d+b*c)^3/(b*x+a)/(d*x+c)^(3/2)+105/8*b^(3/2)*d^3*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c
)^(1/2))/(-a*d+b*c)^(11/2)-105/8*b*d^3/(-a*d+b*c)^5/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {44, 53, 65, 214} \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=\frac {105 b^{3/2} d^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{11/2}}-\frac {105 b d^3}{8 \sqrt {c+d x} (b c-a d)^5}-\frac {35 d^3}{8 (c+d x)^{3/2} (b c-a d)^4}-\frac {21 d^2}{8 (a+b x) (c+d x)^{3/2} (b c-a d)^3}+\frac {3 d}{4 (a+b x)^2 (c+d x)^{3/2} (b c-a d)^2}-\frac {1}{3 (a+b x)^3 (c+d x)^{3/2} (b c-a d)} \]

[In]

Int[1/((a + b*x)^4*(c + d*x)^(5/2)),x]

[Out]

(-35*d^3)/(8*(b*c - a*d)^4*(c + d*x)^(3/2)) - 1/(3*(b*c - a*d)*(a + b*x)^3*(c + d*x)^(3/2)) + (3*d)/(4*(b*c -
a*d)^2*(a + b*x)^2*(c + d*x)^(3/2)) - (21*d^2)/(8*(b*c - a*d)^3*(a + b*x)*(c + d*x)^(3/2)) - (105*b*d^3)/(8*(b
*c - a*d)^5*Sqrt[c + d*x]) + (105*b^(3/2)*d^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*(b*c - a*d)
^(11/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 214

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}-\frac {(3 d) \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx}{2 (b c-a d)} \\ & = -\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}+\frac {\left (21 d^2\right ) \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx}{8 (b c-a d)^2} \\ & = -\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac {21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {\left (105 d^3\right ) \int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx}{16 (b c-a d)^3} \\ & = -\frac {35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac {21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {\left (105 b d^3\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{16 (b c-a d)^4} \\ & = -\frac {35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac {21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {105 b d^3}{8 (b c-a d)^5 \sqrt {c+d x}}-\frac {\left (105 b^2 d^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{16 (b c-a d)^5} \\ & = -\frac {35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac {21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {105 b d^3}{8 (b c-a d)^5 \sqrt {c+d x}}-\frac {\left (105 b^2 d^2\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{8 (b c-a d)^5} \\ & = -\frac {35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac {1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac {3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac {21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {105 b d^3}{8 (b c-a d)^5 \sqrt {c+d x}}+\frac {105 b^{3/2} d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=\frac {1}{24} \left (\frac {-16 a^4 d^4+16 a^3 b d^3 (13 c+9 d x)+3 a^2 b^2 d^2 \left (55 c^2+318 c d x+231 d^2 x^2\right )+2 a b^3 d \left (-25 c^3+90 c^2 d x+567 c d^2 x^2+420 d^3 x^3\right )+b^4 \left (8 c^4-18 c^3 d x+63 c^2 d^2 x^2+420 c d^3 x^3+315 d^4 x^4\right )}{(-b c+a d)^5 (a+b x)^3 (c+d x)^{3/2}}+\frac {315 b^{3/2} d^3 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{11/2}}\right ) \]

[In]

Integrate[1/((a + b*x)^4*(c + d*x)^(5/2)),x]

[Out]

((-16*a^4*d^4 + 16*a^3*b*d^3*(13*c + 9*d*x) + 3*a^2*b^2*d^2*(55*c^2 + 318*c*d*x + 231*d^2*x^2) + 2*a*b^3*d*(-2
5*c^3 + 90*c^2*d*x + 567*c*d^2*x^2 + 420*d^3*x^3) + b^4*(8*c^4 - 18*c^3*d*x + 63*c^2*d^2*x^2 + 420*c*d^3*x^3 +
315*d^4*x^4))/((-(b*c) + a*d)^5*(a + b*x)^3*(c + d*x)^(3/2)) + (315*b^(3/2)*d^3*ArcTan[(Sqrt[b]*Sqrt[c + d*x]
)/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(11/2))/24

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(2 d^{3} \left (-\frac {1}{3 \left (a d -b c \right )^{4} \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b}{\left (a d -b c \right )^{5} \sqrt {d x +c}}+\frac {b^{2} \left (\frac {\frac {41 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{16}+\frac {35 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {3}{2}}}{6}+\left (\frac {55}{16} a^{2} d^{2}-\frac {55}{8} a b c d +\frac {55}{16} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {105 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{5}}\right )\)	\(177\)
default	\(2 d^{3} \left (-\frac {1}{3 \left (a d -b c \right )^{4} \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b}{\left (a d -b c \right )^{5} \sqrt {d x +c}}+\frac {b^{2} \left (\frac {\frac {41 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{16}+\frac {35 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {3}{2}}}{6}+\left (\frac {55}{16} a^{2} d^{2}-\frac {55}{8} a b c d +\frac {55}{16} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {105 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{5}}\right )\)	\(177\)
pseudoelliptic	\(-\frac {2 \left (-\frac {315 b^{2} d^{3} \left (d x +c \right )^{\frac {3}{2}} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16}+\sqrt {\left (a d -b c \right ) b}\, \left (\left (-\frac {315}{16} d^{4} x^{4}-\frac {105}{4} c \,d^{3} x^{3}-\frac {63}{16} c^{2} d^{2} x^{2}+\frac {9}{8} c^{3} d x -\frac {1}{2} c^{4}\right ) b^{4}+\frac {25 \left (-\frac {84}{5} d^{3} x^{3}-\frac {567}{25} c \,d^{2} x^{2}-\frac {18}{5} c^{2} d x +c^{3}\right ) d a \,b^{3}}{8}-\frac {165 d^{2} \left (\frac {21}{5} d^{2} x^{2}+\frac {318}{55} c d x +c^{2}\right ) a^{2} b^{2}}{16}-13 \left (\frac {9 d x}{13}+c \right ) d^{3} a^{3} b +a^{4} d^{4}\right )\right )}{3 \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{3} \left (a d -b c \right )^{5}}\)	\(228\)

[In]

int(1/(b*x+a)^4/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*d^3*(-1/3/(a*d-b*c)^4/(d*x+c)^(3/2)+4/(a*d-b*c)^5*b/(d*x+c)^(1/2)+1/(a*d-b*c)^5*b^2*((41/16*(d*x+c)^(5/2)*b^
2+35/6*(a*d-b*c)*b*(d*x+c)^(3/2)+(55/16*a^2*d^2-55/8*a*b*c*d+55/16*b^2*c^2)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)
^3+105/16/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 915 vs. \(2 (168) = 336\).

Time = 0.35 (sec) , antiderivative size = 1840, normalized size of antiderivative = 9.20 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(b*x+a)^4/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(315*(b^4*d^5*x^5 + a^3*b*c^2*d^3 + (2*b^4*c*d^4 + 3*a*b^3*d^5)*x^4 + (b^4*c^2*d^3 + 6*a*b^3*c*d^4 + 3*
a^2*b^2*d^5)*x^3 + (3*a*b^3*c^2*d^3 + 6*a^2*b^2*c*d^4 + a^3*b*d^5)*x^2 + (3*a^2*b^2*c^2*d^3 + 2*a^3*b*c*d^4)*x
)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d - 2*(b*c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) +
2*(315*b^4*d^4*x^4 + 8*b^4*c^4 - 50*a*b^3*c^3*d + 165*a^2*b^2*c^2*d^2 + 208*a^3*b*c*d^3 - 16*a^4*d^4 + 420*(b
^4*c*d^3 + 2*a*b^3*d^4)*x^3 + 63*(b^4*c^2*d^2 + 18*a*b^3*c*d^3 + 11*a^2*b^2*d^4)*x^2 - 18*(b^4*c^3*d - 10*a*b^
3*c^2*d^2 - 53*a^2*b^2*c*d^3 - 8*a^3*b*d^4)*x)*sqrt(d*x + c))/(a^3*b^5*c^7 - 5*a^4*b^4*c^6*d + 10*a^5*b^3*c^5*
d^2 - 10*a^6*b^2*c^4*d^3 + 5*a^7*b*c^3*d^4 - a^8*c^2*d^5 + (b^8*c^5*d^2 - 5*a*b^7*c^4*d^3 + 10*a^2*b^6*c^3*d^4
- 10*a^3*b^5*c^2*d^5 + 5*a^4*b^4*c*d^6 - a^5*b^3*d^7)*x^5 + (2*b^8*c^6*d - 7*a*b^7*c^5*d^2 + 5*a^2*b^6*c^4*d^
3 + 10*a^3*b^5*c^3*d^4 - 20*a^4*b^4*c^2*d^5 + 13*a^5*b^3*c*d^6 - 3*a^6*b^2*d^7)*x^4 + (b^8*c^7 + a*b^7*c^6*d -
17*a^2*b^6*c^5*d^2 + 35*a^3*b^5*c^4*d^3 - 25*a^4*b^4*c^3*d^4 - a^5*b^3*c^2*d^5 + 9*a^6*b^2*c*d^6 - 3*a^7*b*d^
7)*x^3 + (3*a*b^7*c^7 - 9*a^2*b^6*c^6*d + a^3*b^5*c^5*d^2 + 25*a^4*b^4*c^4*d^3 - 35*a^5*b^3*c^3*d^4 + 17*a^6*b
^2*c^2*d^5 - a^7*b*c*d^6 - a^8*d^7)*x^2 + (3*a^2*b^6*c^7 - 13*a^3*b^5*c^6*d + 20*a^4*b^4*c^5*d^2 - 10*a^5*b^3*
c^4*d^3 - 5*a^6*b^2*c^3*d^4 + 7*a^7*b*c^2*d^5 - 2*a^8*c*d^6)*x), 1/24*(315*(b^4*d^5*x^5 + a^3*b*c^2*d^3 + (2*b
^4*c*d^4 + 3*a*b^3*d^5)*x^4 + (b^4*c^2*d^3 + 6*a*b^3*c*d^4 + 3*a^2*b^2*d^5)*x^3 + (3*a*b^3*c^2*d^3 + 6*a^2*b^2
*c*d^4 + a^3*b*d^5)*x^2 + (3*a^2*b^2*c^2*d^3 + 2*a^3*b*c*d^4)*x)*sqrt(-b/(b*c - a*d))*arctan(-(b*c - a*d)*sqrt
(d*x + c)*sqrt(-b/(b*c - a*d))/(b*d*x + b*c)) - (315*b^4*d^4*x^4 + 8*b^4*c^4 - 50*a*b^3*c^3*d + 165*a^2*b^2*c^
2*d^2 + 208*a^3*b*c*d^3 - 16*a^4*d^4 + 420*(b^4*c*d^3 + 2*a*b^3*d^4)*x^3 + 63*(b^4*c^2*d^2 + 18*a*b^3*c*d^3 +
11*a^2*b^2*d^4)*x^2 - 18*(b^4*c^3*d - 10*a*b^3*c^2*d^2 - 53*a^2*b^2*c*d^3 - 8*a^3*b*d^4)*x)*sqrt(d*x + c))/(a^
3*b^5*c^7 - 5*a^4*b^4*c^6*d + 10*a^5*b^3*c^5*d^2 - 10*a^6*b^2*c^4*d^3 + 5*a^7*b*c^3*d^4 - a^8*c^2*d^5 + (b^8*c
^5*d^2 - 5*a*b^7*c^4*d^3 + 10*a^2*b^6*c^3*d^4 - 10*a^3*b^5*c^2*d^5 + 5*a^4*b^4*c*d^6 - a^5*b^3*d^7)*x^5 + (2*b
^8*c^6*d - 7*a*b^7*c^5*d^2 + 5*a^2*b^6*c^4*d^3 + 10*a^3*b^5*c^3*d^4 - 20*a^4*b^4*c^2*d^5 + 13*a^5*b^3*c*d^6 -
3*a^6*b^2*d^7)*x^4 + (b^8*c^7 + a*b^7*c^6*d - 17*a^2*b^6*c^5*d^2 + 35*a^3*b^5*c^4*d^3 - 25*a^4*b^4*c^3*d^4 - a
^5*b^3*c^2*d^5 + 9*a^6*b^2*c*d^6 - 3*a^7*b*d^7)*x^3 + (3*a*b^7*c^7 - 9*a^2*b^6*c^6*d + a^3*b^5*c^5*d^2 + 25*a^
4*b^4*c^4*d^3 - 35*a^5*b^3*c^3*d^4 + 17*a^6*b^2*c^2*d^5 - a^7*b*c*d^6 - a^8*d^7)*x^2 + (3*a^2*b^6*c^7 - 13*a^3
*b^5*c^6*d + 20*a^4*b^4*c^5*d^2 - 10*a^5*b^3*c^4*d^3 - 5*a^6*b^2*c^3*d^4 + 7*a^7*b*c^2*d^5 - 2*a^8*c*d^6)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x+a)**4/(d*x+c)**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(b*x+a)^4/(d*x+c)^(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 detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (168) = 336\).

Time = 0.33 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=-\frac {105 \, b^{2} d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {-b^{2} c + a b d}} - \frac {315 \, {\left (d x + c\right )}^{4} b^{4} d^{3} - 840 \, {\left (d x + c\right )}^{3} b^{4} c d^{3} + 693 \, {\left (d x + c\right )}^{2} b^{4} c^{2} d^{3} - 144 \, {\left (d x + c\right )} b^{4} c^{3} d^{3} - 16 \, b^{4} c^{4} d^{3} + 840 \, {\left (d x + c\right )}^{3} a b^{3} d^{4} - 1386 \, {\left (d x + c\right )}^{2} a b^{3} c d^{4} + 432 \, {\left (d x + c\right )} a b^{3} c^{2} d^{4} + 64 \, a b^{3} c^{3} d^{4} + 693 \, {\left (d x + c\right )}^{2} a^{2} b^{2} d^{5} - 432 \, {\left (d x + c\right )} a^{2} b^{2} c d^{5} - 96 \, a^{2} b^{2} c^{2} d^{5} + 144 \, {\left (d x + c\right )} a^{3} b d^{6} + 64 \, a^{3} b c d^{6} - 16 \, a^{4} d^{7}}{24 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} {\left ({\left (d x + c\right )}^{\frac {3}{2}} b - \sqrt {d x + c} b c + \sqrt {d x + c} a d\right )}^{3}} \]

[In]

integrate(1/(b*x+a)^4/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-105/8*b^2*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 1
0*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*sqrt(-b^2*c + a*b*d)) - 1/24*(315*(d*x + c)^4*b^4*d^3 - 840*(d*x
+ c)^3*b^4*c*d^3 + 693*(d*x + c)^2*b^4*c^2*d^3 - 144*(d*x + c)*b^4*c^3*d^3 - 16*b^4*c^4*d^3 + 840*(d*x + c)^3*
a*b^3*d^4 - 1386*(d*x + c)^2*a*b^3*c*d^4 + 432*(d*x + c)*a*b^3*c^2*d^4 + 64*a*b^3*c^3*d^4 + 693*(d*x + c)^2*a^
2*b^2*d^5 - 432*(d*x + c)*a^2*b^2*c*d^5 - 96*a^2*b^2*c^2*d^5 + 144*(d*x + c)*a^3*b*d^6 + 64*a^3*b*c*d^6 - 16*a
^4*d^7)/((b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*((d*x +
c)^(3/2)*b - sqrt(d*x + c)*b*c + sqrt(d*x + c)*a*d)^3)

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{5/2}} \, dx=\frac {\frac {231\,b^2\,d^3\,{\left (c+d\,x\right )}^2}{8\,{\left (a\,d-b\,c\right )}^3}-\frac {2\,d^3}{3\,\left (a\,d-b\,c\right )}+\frac {35\,b^3\,d^3\,{\left (c+d\,x\right )}^3}{{\left (a\,d-b\,c\right )}^4}+\frac {105\,b^4\,d^3\,{\left (c+d\,x\right )}^4}{8\,{\left (a\,d-b\,c\right )}^5}+\frac {6\,b\,d^3\,\left (c+d\,x\right )}{{\left (a\,d-b\,c\right )}^2}}{{\left (c+d\,x\right )}^{3/2}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )+b^3\,{\left (c+d\,x\right )}^{9/2}-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{7/2}+{\left (c+d\,x\right )}^{5/2}\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )}+\frac {105\,b^{3/2}\,d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^5\,d^5-5\,a^4\,b\,c\,d^4+10\,a^3\,b^2\,c^2\,d^3-10\,a^2\,b^3\,c^3\,d^2+5\,a\,b^4\,c^4\,d-b^5\,c^5\right )}{{\left (a\,d-b\,c\right )}^{11/2}}\right )}{8\,{\left (a\,d-b\,c\right )}^{11/2}} \]

[In]

int(1/((a + b*x)^4*(c + d*x)^(5/2)),x)

[Out]

((231*b^2*d^3*(c + d*x)^2)/(8*(a*d - b*c)^3) - (2*d^3)/(3*(a*d - b*c)) + (35*b^3*d^3*(c + d*x)^3)/(a*d - b*c)^
4 + (105*b^4*d^3*(c + d*x)^4)/(8*(a*d - b*c)^5) + (6*b*d^3*(c + d*x))/(a*d - b*c)^2)/((c + d*x)^(3/2)*(a^3*d^3
- b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2) + b^3*(c + d*x)^(9/2) - (3*b^3*c - 3*a*b^2*d)*(c + d*x)^(7/2) + (c
+ d*x)^(5/2)*(3*b^3*c^2 + 3*a^2*b*d^2 - 6*a*b^2*c*d)) + (105*b^(3/2)*d^3*atan((b^(1/2)*(c + d*x)^(1/2)*(a^5*d
^5 - b^5*c^5 - 10*a^2*b^3*c^3*d^2 + 10*a^3*b^2*c^2*d^3 + 5*a*b^4*c^4*d - 5*a^4*b*c*d^4))/(a*d - b*c)^(11/2)))/
(8*(a*d - b*c)^(11/2))