\(\int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx\) [1498]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=-\frac {2 \sqrt {c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {4 d \sqrt {c+d x}}{3 (b c-a d)^2 \sqrt {a+b x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, 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} \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\frac {4 d \sqrt {c+d x}}{3 \sqrt {a+b x} (b c-a d)^2}-\frac {2 \sqrt {c+d x}}{3 (a+b x)^{3/2} (b c-a d)} \]

[In]

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

[Out]

(-2*Sqrt[c + d*x])/(3*(b*c - a*d)*(a + b*x)^(3/2)) + (4*d*Sqrt[c + d*x])/(3*(b*c - a*d)^2*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]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=-\frac {2 \sqrt {c+d x} (b c-3 a d-2 b d x)}{3 (b c-a d)^2 (a+b x)^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {2 \sqrt {d x +c}\, \left (2 b d x +3 a d -b c \right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(54\)
default	\(-\frac {2 \sqrt {d x +c}}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 d \sqrt {d x +c}}{3 \left (-a d +b c \right )^{2} \sqrt {b x +a}}\)	\(55\)

[In]

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

[Out]

2/3*(d*x+c)^(1/2)*(2*b*d*x+3*a*d-b*c)/(b*x+a)^(3/2)/(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. 118 vs. \(2 (54) = 108\).

Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\frac {2 \, {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )}} \]

[In]

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

[Out]

2/3*(2*b*d*x - b*c + 3*a*d)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b^4*c^2 - 2*a*
b^3*c*d + a^2*b^2*d^2)*x^2 + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x)

Sympy [F]

\[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(b*x+a)^(5/2)/(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. 121 vs. \(2 (54) = 108\).

Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.83 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\frac {8 \, {\left (b^{2} c - a b d - 3 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} \sqrt {b d} b^{2} d}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} {\left | b \right |}} \]

[In]

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

[Out]

8/3*(b^2*c - a*b*d - 3*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*sqrt(b*d)*b^2*d/((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)^3*abs(b))

Mupad [B] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx=\frac {\left (\frac {4\,d\,x}{3\,{\left (a\,d-b\,c\right )}^2}+\frac {6\,a\,d-2\,b\,c}{3\,b\,{\left (a\,d-b\,c\right )}^2}\right )\,\sqrt {c+d\,x}}{x\,\sqrt {a+b\,x}+\frac {a\,\sqrt {a+b\,x}}{b}} \]

[In]

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

[Out]

(((4*d*x)/(3*(a*d - b*c)^2) + (6*a*d - 2*b*c)/(3*b*(a*d - b*c)^2))*(c + d*x)^(1/2))/(x*(a + b*x)^(1/2) + (a*(a
+ b*x)^(1/2))/b)