\(\int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx\) [780]

Optimal result

Integrand size = 19, antiderivative size = 62 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2}{(b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {4 d \sqrt {a+b x}}{(b c-a d)^2 \sqrt {c+d x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {4 d \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)^2}-\frac {2}{\sqrt {a+b x} \sqrt {c+d x} (b c-a d)} \]

[In]

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

[Out]

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

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{(b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {(2 d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{b c-a d} \\ & = -\frac {2}{(b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {4 d \sqrt {a+b x}}{(b c-a d)^2 \sqrt {c+d x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 (b c+a d+2 b d x)}{(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84

method	result	size
gosper	\(-\frac {2 \left (2 b d x +a d +b c \right )}{\sqrt {b x +a}\, \sqrt {d x +c}\, \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(52\)
default	\(-\frac {2}{\left (-a d +b c \right ) \sqrt {b x +a}\, \sqrt {d x +c}}+\frac {4 d \sqrt {b x +a}}{\left (-a d +b c \right ) \sqrt {d x +c}\, \left (a d -b c \right )}\)	\(65\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (54) = 108\).

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.02 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(1/(b*x+a)^(3/2)/(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. 142 vs. \(2 (54) = 108\).

Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 \, \sqrt {b x + a} b^{2} d}{{\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {4 \, \sqrt {b d} b^{2}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} {\left (b c {\left | b \right |} - a d {\left | b \right |}\right )}} \]

[In]

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

[Out]

-2*sqrt(b*x + a)*b^2*d/((b^2*c^2*abs(b) - 2*a*b*c*d*abs(b) + a^2*d^2*abs(b))*sqrt(b^2*c + (b*x + a)*b*d - a*b*
d)) - 4*sqrt(b*d)*b^2/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*(b*
c*abs(b) - a*d*abs(b)))

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.15 \[ \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {\left (\frac {4\,b\,x}{{\left (a\,d-b\,c\right )}^2}+\frac {2\,a\,d+2\,b\,c}{d\,{\left (a\,d-b\,c\right )}^2}\right )\,\sqrt {c+d\,x}}{x\,\sqrt {a+b\,x}+\frac {c\,\sqrt {a+b\,x}}{d}} \]

[In]

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

[Out]

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