3.1432 \(\int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=140 \[ \frac {15 d^2}{4 \sqrt {c+d x} (b c-a d)^3}-\frac {15 \sqrt {b} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac {5 d}{4 (a+b x) \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)} \]

[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)

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Rubi [A]  time = 0.05, antiderivative size = 140, 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} \[ \frac {15 d^2}{4 \sqrt {c+d x} (b c-a d)^3}-\frac {15 \sqrt {b} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac {5 d}{4 (a+b x) \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[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 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)^3 (c+d x)^{3/2}} \, dx &=-\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) \operatorname {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*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.36 \[ -\frac {2 d^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{\sqrt {c+d x} (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 782, normalized size = 5.59 \[ \left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [B]  time = 1.03, size = 234, normalized size = 1.67 \[ \frac {15 \, b d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, d^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {d x + c}} + \frac {7 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 9 \, \sqrt {d x + c} b^{2} c d^{2} + 9 \, \sqrt {d x + c} a b d^{3}}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 179, normalized size = 1.28 \[ -\frac {9 \sqrt {d x +c}\, a b \,d^{3}}{4 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}+\frac {9 \sqrt {d x +c}\, b^{2} c \,d^{2}}{4 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}-\frac {7 \left (d x +c \right )^{\frac {3}{2}} b^{2} d^{2}}{4 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}-\frac {15 b \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 \left (a d -b c \right )^{3} \sqrt {\left (a d -b c \right ) b}}-\frac {2 d^{2}}{\left (a d -b c \right )^{3} \sqrt {d x +c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*d^2/(a*d-b*c)^3/(d*x+c)^(1/2)-7/4*d^2*b^2/(a*d-b*c)^3/(b*d*x+a*d)^2*(d*x+c)^(3/2)-9/4*d^3*b/(a*d-b*c)^3/(b*
d*x+a*d)^2*(d*x+c)^(1/2)*a+9/4*d^2*b^2/(a*d-b*c)^3/(b*d*x+a*d)^2*(d*x+c)^(1/2)*c-15/4*d^2*b/(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)^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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.44, size = 205, normalized size = 1.46 \[ -\frac {\frac {2\,d^2}{a\,d-b\,c}+\frac {15\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{4\,{\left (a\,d-b\,c\right )}^3}+\frac {25\,b\,d^2\,\left (c+d\,x\right )}{4\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{5/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{3/2}+\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {15\,\sqrt {b}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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