\(\int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx\) [1497]

Optimal result

Integrand size = 19, antiderivative size = 30 \[ \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx=-\frac {2 \sqrt {c+d x}}{(b c-a d) \sqrt {a+b x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx=-\frac {2 \sqrt {c+d x}}{\sqrt {a+b x} (b c-a d)} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {c+d x}}{(b c-a d) \sqrt {a+b x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx=-\frac {2 \sqrt {c+d x}}{(b c-a d) \sqrt {a+b x}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
gosper	\(\frac {2 \sqrt {d x +c}}{\sqrt {b x +a}\, \left (a d -b c \right )}\)	\(27\)
default	\(-\frac {2 \sqrt {d x +c}}{\left (-a d +b c \right ) \sqrt {b x +a}}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx=-\frac {2 \, \sqrt {b x + a} \sqrt {d x + c}}{a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(1/(b*x+a)^(3/2)/(d*x+c)^(1/2),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. 66 vs. \(2 (26) = 52\).

Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.20 \[ \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx=-\frac {4 \, \sqrt {b d} b}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} {\left | b \right |}} \]

[In]

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

[Out]

-4*sqrt(b*d)*b/((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)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx=\frac {2\,\sqrt {c+d\,x}}{\left (a\,d-b\,c\right )\,\sqrt {a+b\,x}} \]

[In]

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

[Out]

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