3.1441 \(\int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=124 \[ \frac {5 b^{3/2} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{7/2}}-\frac {5 b d}{\sqrt {c+d x} (b c-a d)^3}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}-\frac {5 d}{3 (c+d x)^{3/2} (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ \frac {5 b^{3/2} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{7/2}}-\frac {5 b d}{\sqrt {c+d x} (b c-a d)^3}-\frac {1}{(a+b x) (c+d x)^{3/2} (b c-a d)}-\frac {5 d}{3 (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

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 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 208

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.40 \[ -\frac {2 d \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{3 (c+d x)^{3/2} (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*Hypergeometric2F1[-3/2, 2, -1/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^2*(c + d*x)^(3/2))

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fricas [B]  time = 0.48, size = 782, normalized size = 6.31 \[ \left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}}, \frac {15 \, {\left (b^{2} d^{3} x^{3} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - {\left (15 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(15*(b^2*d^3*x^3 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^2 + (b^2*c^2*d + 2*a*b*c*d^2)*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 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x)*sqrt(d*x + c))/(a*b^3*c^5 - 3*a^2*b^2*c^
4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b
^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b
^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x), 1/3*(15*(b^2*d^3*x^3 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^2
+ (b^2*c^2*d + 2*a*b*c*d^2)*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 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x)*sqrt(d*x + c
))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c
*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4
*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)]

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giac [B]  time = 1.11, size = 216, normalized size = 1.74 \[ -\frac {5 \, b^{2} d \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {\sqrt {d x + c} b^{2} d}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} - \frac {2 \, {\left (6 \, {\left (d x + c\right )} b d + b c d - a d^{2}\right )}}{3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*b^2*d*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)*sqr
t(-b^2*c + a*b*d)) - sqrt(d*x + c)*b^2*d/((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)) - 2/3*(6*(d*x + c)*b*d + b*c*d - a*d^2)/((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)^(3/2))

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maple [A]  time = 0.02, size = 125, normalized size = 1.01 \[ \frac {5 b^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3} \sqrt {\left (a d -b c \right ) b}}+\frac {\sqrt {d x +c}\, b^{2} d}{\left (a d -b c \right )^{3} \left (b d x +a d \right )}+\frac {4 b d}{\left (a d -b c \right )^{3} \sqrt {d x +c}}-\frac {2 d}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(b*x+a)^2/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.38, size = 161, normalized size = 1.30 \[ \frac {\frac {10\,b\,d\,\left (c+d\,x\right )}{3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,d}{3\,\left (a\,d-b\,c\right )}+\frac {5\,b^2\,d\,{\left (c+d\,x\right )}^2}{{\left (a\,d-b\,c\right )}^3}}{b\,{\left (c+d\,x\right )}^{5/2}+\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}+\frac {5\,b^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{{\left (a\,d-b\,c\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((10*b*d*(c + d*x))/(3*(a*d - b*c)^2) - (2*d)/(3*(a*d - b*c)) + (5*b^2*d*(c + d*x)^2)/(a*d - b*c)^3)/(b*(c + d
*x)^(5/2) + (a*d - b*c)*(c + d*x)^(3/2)) + (5*b^(3/2)*d*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)))/(a*d - b*c)^(7/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{2} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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