\(\int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx\) [1442]

Optimal result

Integrand size = 17, antiderivative size = 167 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}-\frac {35 b^{3/2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}} \]

[Out]

35/12*d^2/(-a*d+b*c)^3/(d*x+c)^(3/2)-1/2/(-a*d+b*c)/(b*x+a)^2/(d*x+c)^(3/2)+7/4*d/(-a*d+b*c)^2/(b*x+a)/(d*x+c)
^(3/2)-35/4*b^(3/2)*d^2*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(9/2)+35/4*b*d^2/(-a*d+b*c)
^4/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {44, 53, 65, 214} \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=-\frac {35 b^{3/2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}+\frac {35 b d^2}{4 \sqrt {c+d x} (b c-a d)^4}+\frac {35 d^2}{12 (c+d x)^{3/2} (b c-a d)^3}+\frac {7 d}{4 (a+b x) (c+d x)^{3/2} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 (c+d x)^{3/2} (b c-a d)} \]

[In]

Int[1/((a + b*x)^3*(c + d*x)^(5/2)),x]

[Out]

(35*d^2)/(12*(b*c - a*d)^3*(c + d*x)^(3/2)) - 1/(2*(b*c - a*d)*(a + b*x)^2*(c + d*x)^(3/2)) + (7*d)/(4*(b*c -
a*d)^2*(a + b*x)*(c + d*x)^(3/2)) + (35*b*d^2)/(4*(b*c - a*d)^4*Sqrt[c + d*x]) - (35*b^(3/2)*d^2*ArcTanh[(Sqrt
[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*(b*c - a*d)^(9/2))

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 (c+d x)^{3/2}}-\frac {(7 d) \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx}{4 (b c-a d)} \\ & = -\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {\left (35 d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx}{8 (b c-a d)^2} \\ & = \frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {\left (35 b d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{8 (b c-a d)^3} \\ & = \frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}+\frac {\left (35 b^2 d^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 (b c-a d)^4} \\ & = \frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}+\frac {\left (35 b^2 d\right ) \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)^4} \\ & = \frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}-\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\frac {-8 a^3 d^3+8 a^2 b d^2 (10 c+7 d x)+a b^2 d \left (39 c^2+238 c d x+175 d^2 x^2\right )+b^3 \left (-6 c^3+21 c^2 d x+140 c d^2 x^2+105 d^3 x^3\right )}{12 (b c-a d)^4 (a+b x)^2 (c+d x)^{3/2}}+\frac {35 b^{3/2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{4 (-b c+a d)^{9/2}} \]

[In]

Integrate[1/((a + b*x)^3*(c + d*x)^(5/2)),x]

[Out]

(-8*a^3*d^3 + 8*a^2*b*d^2*(10*c + 7*d*x) + a*b^2*d*(39*c^2 + 238*c*d*x + 175*d^2*x^2) + b^3*(-6*c^3 + 21*c^2*d
*x + 140*c*d^2*x^2 + 105*d^3*x^3))/(12*(b*c - a*d)^4*(a + b*x)^2*(c + d*x)^(3/2)) + (35*b^(3/2)*d^2*ArcTan[(Sq
rt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(4*(-(b*c) + a*d)^(9/2))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(2 d^{2} \left (-\frac {1}{3 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {3}{2}}}+\frac {3 b}{\left (a d -b c \right )^{4} \sqrt {d x +c}}+\frac {b^{2} \left (\frac {\frac {11 b \left (d x +c \right )^{\frac {3}{2}}}{8}+\left (\frac {13 a d}{8}-\frac {13 b c}{8}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{4}}\right )\)	\(143\)
default	\(2 d^{2} \left (-\frac {1}{3 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {3}{2}}}+\frac {3 b}{\left (a d -b c \right )^{4} \sqrt {d x +c}}+\frac {b^{2} \left (\frac {\frac {11 b \left (d x +c \right )^{\frac {3}{2}}}{8}+\left (\frac {13 a d}{8}-\frac {13 b c}{8}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{4}}\right )\)	\(143\)
pseudoelliptic	\(-\frac {2 \left (-\frac {105 \left (d x +c \right )^{\frac {3}{2}} b^{2} d^{2} \left (b x +a \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8}+\sqrt {\left (a d -b c \right ) b}\, \left (\left (-\frac {105}{8} d^{3} x^{3}-\frac {35}{2} c \,d^{2} x^{2}-\frac {21}{8} c^{2} d x +\frac {3}{4} c^{3}\right ) b^{3}-\frac {39 \left (\frac {175}{39} d^{2} x^{2}+\frac {238}{39} c d x +c^{2}\right ) d a \,b^{2}}{8}-10 \left (\frac {7 d x}{10}+c \right ) d^{2} a^{2} b +a^{3} d^{3}\right )\right )}{3 \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{2} \left (a d -b c \right )^{4}}\)	\(178\)

[In]

int(1/(b*x+a)^3/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*d^2*(-1/3/(a*d-b*c)^3/(d*x+c)^(3/2)+3/(a*d-b*c)^4*b/(d*x+c)^(1/2)+1/(a*d-b*c)^4*b^2*((11/8*b*(d*x+c)^(3/2)+(
13/8*a*d-13/8*b*c)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)^2+35/8/((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. 608 vs. \(2 (139) = 278\).

Time = 0.27 (sec) , antiderivative size = 1226, normalized size of antiderivative = 7.34 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*d^4*x^4 + a^2*b*c^2*d^2 + 2*(b^3*c*d^3 + a*b^2*d^4)*x^3 + (b^3*c^2*d^2 + 4*a*b^2*c*d^3 + a^2*b
*d^4)*x^2 + 2*(a*b^2*c^2*d^2 + a^2*b*c*d^3)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d - 2*(b*c - a*d)*sq
rt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) + 2*(105*b^3*d^3*x^3 - 6*b^3*c^3 + 39*a*b^2*c^2*d + 80*a^2*b*c*d^2
- 8*a^3*d^3 + 35*(4*b^3*c*d^2 + 5*a*b^2*d^3)*x^2 + 7*(3*b^3*c^2*d + 34*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x
+ c))/(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a
*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2
*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^
3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*
c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x), -1/12*(105*(b^3*d^4*x^4 + a^2*b*c^2*d^2 + 2*(b^3*c*d^3 + a*b^2*d^4)
*x^3 + (b^3*c^2*d^2 + 4*a*b^2*c*d^3 + a^2*b*d^4)*x^2 + 2*(a*b^2*c^2*d^2 + a^2*b*c*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)) - (105*b^3*d^3*x^3 - 6*b^3*c^3 + 39*a*b
^2*c^2*d + 80*a^2*b*c*d^2 - 8*a^3*d^3 + 35*(4*b^3*c*d^2 + 5*a*b^2*d^3)*x^2 + 7*(3*b^3*c^2*d + 34*a*b^2*c*d^2 +
8*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2
*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d
- 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^
2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3
*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x+a)**3/(d*x+c)**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(b*x+a)^3/(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. 298 vs. \(2 (139) = 278\).

Time = 0.34 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.78 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\frac {35 \, b^{2} d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (9 \, {\left (d x + c\right )} b d^{2} + b c d^{2} - a d^{3}\right )}}{3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} d^{2} - 13 \, \sqrt {d x + c} b^{3} c d^{2} + 13 \, \sqrt {d x + c} a b^{2} d^{3}}{4 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

35/4*b^2*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^
3*b*c*d^3 + a^4*d^4)*sqrt(-b^2*c + a*b*d)) + 2/3*(9*(d*x + c)*b*d^2 + b*c*d^2 - a*d^3)/((b^4*c^4 - 4*a*b^3*c^3
*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*(d*x + c)^(3/2)) + 1/4*(11*(d*x + c)^(3/2)*b^3*d^2 - 13*sqrt
(d*x + c)*b^3*c*d^2 + 13*sqrt(d*x + c)*a*b^2*d^3)/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^
3 + a^4*d^4)*((d*x + c)*b - b*c + a*d)^2)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx=\frac {\frac {175\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{12\,{\left (a\,d-b\,c\right )}^3}-\frac {2\,d^2}{3\,\left (a\,d-b\,c\right )}+\frac {35\,b^3\,d^2\,{\left (c+d\,x\right )}^3}{4\,{\left (a\,d-b\,c\right )}^4}+\frac {14\,b\,d^2\,\left (c+d\,x\right )}{3\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{7/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{5/2}+{\left (c+d\,x\right )}^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {35\,b^{3/2}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^4\,d^4-4\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d+b^4\,c^4\right )}{{\left (a\,d-b\,c\right )}^{9/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{9/2}} \]

[In]

int(1/((a + b*x)^3*(c + d*x)^(5/2)),x)

[Out]

((175*b^2*d^2*(c + d*x)^2)/(12*(a*d - b*c)^3) - (2*d^2)/(3*(a*d - b*c)) + (35*b^3*d^2*(c + d*x)^3)/(4*(a*d - b
*c)^4) + (14*b*d^2*(c + d*x))/(3*(a*d - b*c)^2))/(b^2*(c + d*x)^(7/2) - (2*b^2*c - 2*a*b*d)*(c + d*x)^(5/2) +
(c + d*x)^(3/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (35*b^(3/2)*d^2*atan((b^(1/2)*(c + d*x)^(1/2)*(a^4*d^4 + b^
4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3))/(a*d - b*c)^(9/2)))/(4*(a*d - b*c)^(9/2))