3.14.61 \(\int \frac {\sqrt {c+d x}}{(a+b x)^{7/2}} \, dx\)

Optimal. Leaf size=66 \[ \frac {4 d (c+d x)^{3/2}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac {2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)} \]

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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 (c+d x)^{3/2}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac {2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x)^(7/2),x]

[Out]

(-2*(c + d*x)^(3/2))/(5*(b*c - a*d)*(a + b*x)^(5/2)) + (4*d*(c + d*x)^(3/2))/(15*(b*c - a*d)^2*(a + b*x)^(3/2)
)

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 {\sqrt {c+d x}}{(a+b x)^{7/2}} \, dx &=-\frac {2 (c+d x)^{3/2}}{5 (b c-a d) (a+b x)^{5/2}}-\frac {(2 d) \int \frac {\sqrt {c+d x}}{(a+b x)^{5/2}} \, dx}{5 (b c-a d)}\\ &=-\frac {2 (c+d x)^{3/2}}{5 (b c-a d) (a+b x)^{5/2}}+\frac {4 d (c+d x)^{3/2}}{15 (b c-a d)^2 (a+b x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.70 \begin {gather*} \frac {2 (c+d x)^{3/2} (5 a d-3 b c+2 b d x)}{15 (a+b x)^{5/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]/(a + b*x)^(7/2),x]

[Out]

(2*(c + d*x)^(3/2)*(-3*b*c + 5*a*d + 2*b*d*x))/(15*(b*c - a*d)^2*(a + b*x)^(5/2))

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IntegrateAlgebraic [A]  time = 0.10, size = 57, normalized size = 0.86 \begin {gather*} -\frac {2 \left (\frac {3 b (c+d x)^{5/2}}{(a+b x)^{5/2}}-\frac {5 d (c+d x)^{3/2}}{(a+b x)^{3/2}}\right )}{15 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[c + d*x]/(a + b*x)^(7/2),x]

[Out]

(-2*((-5*d*(c + d*x)^(3/2))/(a + b*x)^(3/2) + (3*b*(c + d*x)^(5/2))/(a + b*x)^(5/2)))/(15*(b*c - a*d)^2)

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fricas [B]  time = 2.32, size = 175, normalized size = 2.65 \begin {gather*} \frac {2 \, {\left (2 \, b d^{2} x^{2} - 3 \, b c^{2} + 5 \, a c d - {\left (b c d - 5 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{15 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3} + 3 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)/(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

2/15*(2*b*d^2*x^2 - 3*b*c^2 + 5*a*c*d - (b*c*d - 5*a*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^3*b^2*c^2 - 2*a^4*
b*c*d + a^5*d^2 + (b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*x^3 + 3*(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2)*x^2
+ 3*(a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*x)

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giac [B]  time = 1.44, size = 447, normalized size = 6.77 \begin {gather*} \frac {8 \, {\left (\sqrt {b d} b^{7} c^{3} d^{2} - 3 \, \sqrt {b d} a b^{6} c^{2} d^{3} + 3 \, \sqrt {b d} a^{2} b^{5} c d^{4} - \sqrt {b d} a^{3} b^{4} d^{5} - 5 \, \sqrt {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} b^{5} c^{2} d^{2} + 10 \, \sqrt {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} a b^{4} c d^{3} - 5 \, \sqrt {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} a^{2} b^{3} d^{4} - 5 \, \sqrt {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 )}^{4} b^{3} c d^{2} + 5 \, \sqrt {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 )}^{4} a b^{2} d^{3} - 15 \, \sqrt {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 )}^{6} b d^{2}\right )} {\left | b \right |}}{15 \, {\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 )}^{5} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)/(b*x+a)^(7/2),x, algorithm="giac")

[Out]

8/15*(sqrt(b*d)*b^7*c^3*d^2 - 3*sqrt(b*d)*a*b^6*c^2*d^3 + 3*sqrt(b*d)*a^2*b^5*c*d^4 - sqrt(b*d)*a^3*b^4*d^5 -
5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^5*c^2*d^2 + 10*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^4*c*d^3 - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^3*d^4 - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^4*b^3*c*d^2 + 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^4*a*b^2*d^3 - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b*d^2)*abs(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)^5*b^2)

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maple [A]  time = 0.01, size = 54, normalized size = 0.82 \begin {gather*} \frac {2 \left (d x +c \right )^{\frac {3}{2}} \left (2 b d x +5 a d -3 b c \right )}{15 \left (b x +a \right )^{\frac {5}{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((d*x+c)^(1/2)/(b*x+a)^(7/2),x)

[Out]

2/15*(d*x+c)^(3/2)*(2*b*d*x+5*a*d-3*b*c)/(b*x+a)^(5/2)/(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((d*x+c)^(1/2)/(b*x+a)^(7/2),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 = 0.82, size = 127, normalized size = 1.92 \begin {gather*} \frac {\sqrt {c+d\,x}\,\left (\frac {x\,\left (10\,a\,d^2-2\,b\,c\,d\right )}{15\,b^2\,{\left (a\,d-b\,c\right )}^2}-\frac {6\,b\,c^2-10\,a\,c\,d}{15\,b^2\,{\left (a\,d-b\,c\right )}^2}+\frac {4\,d^2\,x^2}{15\,b\,{\left (a\,d-b\,c\right )}^2}\right )}{x^2\,\sqrt {a+b\,x}+\frac {a^2\,\sqrt {a+b\,x}}{b^2}+\frac {2\,a\,x\,\sqrt {a+b\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/2)/(a + b*x)^(7/2),x)

[Out]

((c + d*x)^(1/2)*((x*(10*a*d^2 - 2*b*c*d))/(15*b^2*(a*d - b*c)^2) - (6*b*c^2 - 10*a*c*d)/(15*b^2*(a*d - b*c)^2
) + (4*d^2*x^2)/(15*b*(a*d - b*c)^2)))/(x^2*(a + b*x)^(1/2) + (a^2*(a + b*x)^(1/2))/b^2 + (2*a*x*(a + b*x)^(1/
2))/b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(1/2)/(b*x+a)**(7/2),x)

[Out]

Timed out

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