\(\int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx\) [744]

Optimal result

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

[Out]

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

[In]

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

[Out]

(2*Sqrt[a + b*x])/((b*c - a*d)*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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {2 \sqrt {a+b x}}{(b c-a d) \sqrt {c+d x}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

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

[In]

int(1/(b*x+a)^(1/2)/(d*x+c)^(3/2),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 = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x + a} \sqrt {d x + c}}{b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(1/(b*x+a)^(1/2)/(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 [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x + a} b^{2}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (b c {\left | b \right |} - a d {\left | b \right |}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=-\frac {2\,\sqrt {a+b\,x}}{\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}} \]

[In]

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

[Out]

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