3.1467 \(\int \frac{\sqrt{c+d x}}{(a+b x)^{7/2}} \, dx\)

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

[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)
)

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Rubi [A]  time = 0.0086535, antiderivative size = 66, normalized size of antiderivative = 1., 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} \[ \frac{4 d (c+d x)^{3/2}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac{2 (c+d x)^{3/2}}{5 (a+b x)^{5/2} (b c-a d)} \]

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 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])

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*} \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*}

Mathematica [A]  time = 0.0201384, size = 46, normalized size = 0.7 \[ \frac{2 (c+d x)^{3/2} (5 a d-3 b c+2 b d x)}{15 (a+b x)^{5/2} (b c-a d)^2} \]

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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Maple [A]  time = 0.005, size = 54, normalized size = 0.8 \begin{align*}{\frac{4\,bdx+10\,ad-6\,bc}{15\,{a}^{2}{d}^{2}-30\,abcd+15\,{b}^{2}{c}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}} \left ( bx+a \right ) ^{-{\frac{5}{2}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 4.31271, size = 365, normalized size = 5.53 \begin{align*} \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{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [B]  time = 1.22328, size = 603, normalized size = 9.14 \begin{align*} \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{align*}

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)