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

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

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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} \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[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 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}{(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 {(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*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.70 \begin {gather*} \frac {2 \sqrt {c+d x} (3 a d-b c+2 b d x)}{3 (a+b x)^{3/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[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))

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IntegrateAlgebraic [A]  time = 0.10, size = 56, normalized size = 0.85 \begin {gather*} -\frac {2 \left (\frac {b (c+d x)^{3/2}}{(a+b x)^{3/2}}-\frac {3 d \sqrt {c+d x}}{\sqrt {a+b x}}\right )}{3 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.92, size = 118, normalized size = 1.79 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 1.08, size = 121, normalized size = 1.83 \begin {gather*} \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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.00, size = 54, normalized size = 0.82 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 0.89, size = 71, normalized size = 1.08 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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