\(\int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx\) [1432]

Optimal result

Integrand size = 17, antiderivative size = 140 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}-\frac {15 \sqrt {b} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}} \]

[Out]

-15/4*d^2*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))*b^(1/2)/(-a*d+b*c)^(7/2)+15/4*d^2/(-a*d+b*c)^3/(d*x+
c)^(1/2)-1/2/(-a*d+b*c)/(b*x+a)^2/(d*x+c)^(1/2)+5/4*d/(-a*d+b*c)^2/(b*x+a)/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {44, 53, 65, 214} \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {15 \sqrt {b} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac {15 d^2}{4 \sqrt {c+d x} (b c-a d)^3}+\frac {5 d}{4 (a+b x) \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)} \]

[In]

Int[1/((a + b*x)^3*(c + d*x)^(3/2)),x]

[Out]

(15*d^2)/(4*(b*c - a*d)^3*Sqrt[c + d*x]) - 1/(2*(b*c - a*d)*(a + b*x)^2*Sqrt[c + d*x]) + (5*d)/(4*(b*c - a*d)^
2*(a + b*x)*Sqrt[c + d*x]) - (15*Sqrt[b]*d^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*(b*c - a*d)^
(7/2))

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}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}-\frac {(5 d) \int \frac {1}{(a+b x)^2 (c+d x)^{3/2}} \, dx}{4 (b c-a d)} \\ & = -\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {\left (15 d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{8 (b c-a d)^2} \\ & = \frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {\left (15 b d^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 (b c-a d)^3} \\ & = \frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {(15 b d) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 (b c-a d)^3} \\ & = \frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}-\frac {15 \sqrt {b} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {1}{4} \left (\frac {8 a^2 d^2+a b d (9 c+25 d x)+b^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )}{(b c-a d)^3 (a+b x)^2 \sqrt {c+d x}}-\frac {15 \sqrt {b} d^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}}\right ) \]

[In]

Integrate[1/((a + b*x)^3*(c + d*x)^(3/2)),x]

[Out]

((8*a^2*d^2 + a*b*d*(9*c + 25*d*x) + b^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2))/((b*c - a*d)^3*(a + b*x)^2*Sqrt[c +
d*x]) - (15*Sqrt[b]*d^2*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2))/4

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(2 d^{2} \left (-\frac {1}{\left (a d -b c \right )^{3} \sqrt {d x +c}}-\frac {b \left (\frac {\frac {7 b \left (d x +c \right )^{\frac {3}{2}}}{8}+\left (\frac {9 a d}{8}-\frac {9 b c}{8}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\)	\(122\)
default	\(2 d^{2} \left (-\frac {1}{\left (a d -b c \right )^{3} \sqrt {d x +c}}-\frac {b \left (\frac {\frac {7 b \left (d x +c \right )^{\frac {3}{2}}}{8}+\left (\frac {9 a d}{8}-\frac {9 b c}{8}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\)	\(122\)
pseudoelliptic	\(-\frac {15 \left (\sqrt {d x +c}\, b \,d^{2} \left (b x +a \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\frac {8 \left (\left (\frac {15}{8} d^{2} x^{2}+\frac {5}{8} c d x -\frac {1}{4} c^{2}\right ) b^{2}+\frac {9 d a \left (\frac {25 d x}{9}+c \right ) b}{8}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}}{15}\right )}{4 \sqrt {d x +c}\, \sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{2} \left (a d -b c \right )^{3}}\)	\(137\)

[In]

int(1/(b*x+a)^3/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*d^2*(-1/(a*d-b*c)^3/(d*x+c)^(1/2)-1/(a*d-b*c)^3*b*((7/8*b*(d*x+c)^(3/2)+(9/8*a*d-9/8*b*c)*(d*x+c)^(1/2))/((d
*x+c)*b+a*d-b*c)^2+15/8/((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. 386 vs. \(2 (116) = 232\).

Time = 0.25 (sec) , antiderivative size = 782, normalized size of antiderivative = 5.59 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}\right ] \]

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(15*(b^2*d^3*x^3 + a^2*c*d^2 + (b^2*c*d^2 + 2*a*b*d^3)*x^2 + (2*a*b*c*d^2 + a^2*d^3)*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*(15*b^2*d^2*x^2
- 2*b^2*c^2 + 9*a*b*c*d + 8*a^2*d^2 + 5*(b^2*c*d + 5*a*b*d^2)*x)*sqrt(d*x + c))/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*
d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4
- a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a
^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x), -1/4*(15*(b^2*d^3*x^3 + a^2*c*d^2 + (b^2*c*d^2 + 2*a*b*d^3)*x^2 +
(2*a*b*c*d^2 + a^2*d^3)*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)) - (15*b^2*d^2*x^2 - 2*b^2*c^2 + 9*a*b*c*d + 8*a^2*d^2 + 5*(b^2*c*d + 5*a*b*d^2)*x)*sqrt(d*x + c))/(a^2
*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^
3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^
4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x+a)**3/(d*x+c)**(3/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (116) = 232\).

Time = 0.34 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {15 \, b d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, d^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {d x + c}} + \frac {7 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 9 \, \sqrt {d x + c} b^{2} c d^{2} + 9 \, \sqrt {d x + c} a b d^{3}}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

15/4*b*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*s
qrt(-b^2*c + a*b*d)) + 2*d^2/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(d*x + c)) + 1/4*(7*(d*x
+ c)^(3/2)*b^2*d^2 - 9*sqrt(d*x + c)*b^2*c*d^2 + 9*sqrt(d*x + c)*a*b*d^3)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b
*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c + a*d)^2)

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {\frac {2\,d^2}{a\,d-b\,c}+\frac {15\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{4\,{\left (a\,d-b\,c\right )}^3}+\frac {25\,b\,d^2\,\left (c+d\,x\right )}{4\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{5/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{3/2}+\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {15\,\sqrt {b}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{7/2}} \]

[In]

int(1/((a + b*x)^3*(c + d*x)^(3/2)),x)

[Out]

- ((2*d^2)/(a*d - b*c) + (15*b^2*d^2*(c + d*x)^2)/(4*(a*d - b*c)^3) + (25*b*d^2*(c + d*x))/(4*(a*d - b*c)^2))/
(b^2*(c + d*x)^(5/2) - (2*b^2*c - 2*a*b*d)*(c + d*x)^(3/2) + (c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))
- (15*b^(1/2)*d^2*atan((b^(1/2)*(c + d*x)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(a*d - b*
c)^(7/2)))/(4*(a*d - b*c)^(7/2))