3.15.40 \(\int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx\) [1440]

Optimal. Leaf size=93 \[ \frac {2}{3 (b c-a d) (c+d x)^{3/2}}+\frac {2 b}{(b c-a d)^2 \sqrt {c+d x}}-\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {53, 65, 214} \begin {gather*} -\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac {2 b}{\sqrt {c+d x} (b c-a d)^2}+\frac {2}{3 (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(5/2)),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.19, size = 85, normalized size = 0.91 \begin {gather*} \frac {2 (4 b c-a d+3 b d x)}{3 (b c-a d)^2 (c+d x)^{3/2}}+\frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.17, size = 90, normalized size = 0.97

method	result	size
derivativedivides	\(-\frac {2}{3 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{2} \sqrt {d x +c}}+\frac {2 b^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}\)	\(90\)
default	\(-\frac {2}{3 \left (a d -b c \right ) \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{2} \sqrt {d x +c}}+\frac {2 b^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(a*d-b*c)/(d*x+c)^(3/2)+2/(a*d-b*c)^2*b/(d*x+c)^(1/2)+2*b^2/(a*d-b*c)^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x
+c)^(1/2)/((a*d-b*c)*b)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (77) = 154\).
time = 0.43, size = 398, normalized size = 4.28 \begin {gather*} \left [\frac {3 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\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 (3 \, b d x + 4 \, b c - a d\right )} \sqrt {d x + c}}{3 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x\right )}}, -\frac {2 \, {\left (3 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\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 (3 \, b d x + 4 \, b c - a d\right )} \sqrt {d x + c}\right )}}{3 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 5.97, size = 83, normalized size = 0.89 \begin {gather*} \frac {2 b}{\sqrt {c + d x} \left (a d - b c\right )^{2}} + \frac {2 b \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{\sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )^{2}} - \frac {2}{3 \left (c + d x\right )^{\frac {3}{2}} \left (a d - b c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b/(sqrt(c + d*x)*(a*d - b*c)**2) + 2*b*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(sqrt((a*d - b*c)/b)*(a*d - b
*c)**2) - 2/(3*(c + d*x)**(3/2)*(a*d - b*c))

________________________________________________________________________________________

Giac [A]
time = 2.67, size = 113, normalized size = 1.22 \begin {gather*} \frac {2 \, b^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )} b + b c - a d\right )}}{3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-b^2*c + a*b*d)) + 2/
3*(3*(d*x + c)*b + b*c - a*d)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x + c)^(3/2))

________________________________________________________________________________________

Mupad [B]
time = 0.33, size = 100, normalized size = 1.08 \begin {gather*} \frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {2}{3\,\left (a\,d-b\,c\right )}-\frac {2\,b\,\left (c+d\,x\right )}{{\left (a\,d-b\,c\right )}^2}}{{\left (c+d\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)*(c + d*x)^(5/2)),x)

[Out]

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

________________________________________________________________________________________