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

Optimal. Leaf size=66 \[ \frac {4 b \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac {4 b \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x])/(3*(b*c - a*d)*(c + d*x)^(3/2)) + (4*b*Sqrt[a + b*x])/(3*(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 45

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*} \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx &=\frac {2 \sqrt {a+b x}}{3 (b c-a d) (c+d x)^{3/2}}+\frac {(2 b) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 (b c-a d)}\\ &=\frac {2 \sqrt {a+b x}}{3 (b c-a d) (c+d x)^{3/2}}+\frac {4 b \sqrt {a+b x}}{3 (b c-a d)^2 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.70 \[ \frac {2 \sqrt {a+b x} (-a d+3 b c+2 b d x)}{3 (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.02, size = 118, normalized size = 1.79 \[ \frac {2 \, {\left (2 \, b d x + 3 \, b c - a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.12, size = 126, normalized size = 1.91 \[ \frac {2 \, {\left (\frac {2 \, {\left (b x + a\right )} b^{4} d^{2}}{b^{2} c^{2} d {\left | b \right |} - 2 \, a b c d^{2} {\left | b \right |} + a^{2} d^{3} {\left | b \right |}} + \frac {3 \, {\left (b^{5} c d - a b^{4} d^{2}\right )}}{b^{2} c^{2} d {\left | b \right |} - 2 \, a b c d^{2} {\left | b \right |} + a^{2} d^{3} {\left | b \right |}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(2*(b*x + a)*b^4*d^2/(b^2*c^2*d*abs(b) - 2*a*b*c*d^2*abs(b) + a^2*d^3*abs(b)) + 3*(b^5*c*d - a*b^4*d^2)/(b
^2*c^2*d*abs(b) - 2*a*b*c*d^2*abs(b) + a^2*d^3*abs(b)))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2)

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maple [A]  time = 0.01, size = 53, normalized size = 0.80 \[ -\frac {2 \sqrt {b x +a}\, \left (-2 b d x +a d -3 b c \right )}{3 \left (d x +c \right )^{\frac {3}{2}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(d*x+c)^(5/2),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 zero or nonzero?

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mupad [B]  time = 1.44, size = 127, normalized size = 1.92 \[ \frac {\sqrt {c+d\,x}\,\left (\frac {x\,\left (6\,c\,b^2+2\,a\,d\,b\right )}{3\,d^2\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,a^2\,d-6\,a\,b\,c}{3\,d^2\,{\left (a\,d-b\,c\right )}^2}+\frac {4\,b^2\,x^2}{3\,d\,{\left (a\,d-b\,c\right )}^2}\right )}{x^2\,\sqrt {a+b\,x}+\frac {c^2\,\sqrt {a+b\,x}}{d^2}+\frac {2\,c\,x\,\sqrt {a+b\,x}}{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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