3.1519 \(\int \frac{1}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{32 b d^2 \sqrt{a+b x}}{3 \sqrt{c+d x} (b c-a d)^4}+\frac{16 d^2 \sqrt{a+b x}}{3 (c+d x)^{3/2} (b c-a d)^3}+\frac{4 d}{\sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^2}-\frac{2}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)} \]

[Out]

-2/(3*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) + (4*d)/((b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(3/2)) + (16
*d^2*Sqrt[a + b*x])/(3*(b*c - a*d)^3*(c + d*x)^(3/2)) + (32*b*d^2*Sqrt[a + b*x])/(3*(b*c - a*d)^4*Sqrt[c + d*x
])

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Rubi [A]  time = 0.0284987, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{32 b d^2 \sqrt{a+b x}}{3 \sqrt{c+d x} (b c-a d)^4}+\frac{16 d^2 \sqrt{a+b x}}{3 (c+d x)^{3/2} (b c-a d)^3}+\frac{4 d}{\sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^2}-\frac{2}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(3*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) + (4*d)/((b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(3/2)) + (16
*d^2*Sqrt[a + b*x])/(3*(b*c - a*d)^3*(c + d*x)^(3/2)) + (32*b*d^2*Sqrt[a + b*x])/(3*(b*c - a*d)^4*Sqrt[c + d*x
])

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{1}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx &=-\frac{2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}-\frac{(2 d) \int \frac{1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx}{b c-a d}\\ &=-\frac{2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac{4 d}{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}+\frac{\left (8 d^2\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{5/2}} \, dx}{(b c-a d)^2}\\ &=-\frac{2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac{4 d}{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}+\frac{16 d^2 \sqrt{a+b x}}{3 (b c-a d)^3 (c+d x)^{3/2}}+\frac{\left (16 b d^2\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 (b c-a d)^3}\\ &=-\frac{2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac{4 d}{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}+\frac{16 d^2 \sqrt{a+b x}}{3 (b c-a d)^3 (c+d x)^{3/2}}+\frac{32 b d^2 \sqrt{a+b x}}{3 (b c-a d)^4 \sqrt{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.0506458, size = 118, normalized size = 0.87 \[ \frac{6 a^2 b d^2 (3 c+2 d x)-2 a^3 d^3+6 a b^2 d \left (3 c^2+12 c d x+8 d^2 x^2\right )+b^3 \left (12 c^2 d x-2 c^3+48 c d^2 x^2+32 d^3 x^3\right )}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*d^3 + 6*a^2*b*d^2*(3*c + 2*d*x) + 6*a*b^2*d*(3*c^2 + 12*c*d*x + 8*d^2*x^2) + b^3*(-2*c^3 + 12*c^2*d*x
+ 48*c*d^2*x^2 + 32*d^3*x^3))/(3*(b*c - a*d)^4*(a + b*x)^(3/2)*(c + d*x)^(3/2))

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Maple [A]  time = 0.007, size = 169, normalized size = 1.3 \begin{align*} -{\frac{-32\,{b}^{3}{d}^{3}{x}^{3}-48\,a{b}^{2}{d}^{3}{x}^{2}-48\,{b}^{3}c{d}^{2}{x}^{2}-12\,{a}^{2}b{d}^{3}x-72\,a{b}^{2}c{d}^{2}x-12\,{b}^{3}{c}^{2}dx+2\,{a}^{3}{d}^{3}-18\,{a}^{2}bc{d}^{2}-18\,a{b}^{2}{c}^{2}d+2\,{b}^{3}{c}^{3}}{3\,{d}^{4}{a}^{4}-12\,b{d}^{3}c{a}^{3}+18\,{b}^{2}{d}^{2}{c}^{2}{a}^{2}-12\,{b}^{3}d{c}^{3}a+3\,{b}^{4}{c}^{4}} \left ( bx+a \right ) ^{-{\frac{3}{2}}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-16*b^3*d^3*x^3-24*a*b^2*d^3*x^2-24*b^3*c*d^2*x^2-6*a^2*b*d^3*x-36*a*b^2*c*d^2*x-6*b^3*c^2*d*x+a^3*d^3-9
*a^2*b*c*d^2-9*a*b^2*c^2*d+b^3*c^3)/(b*x+a)^(3/2)/(d*x+c)^(3/2)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d^2-4*a*b
^3*c^3*d+b^4*c^4)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 11.7611, size = 886, normalized size = 6.56 \begin{align*} \frac{2 \,{\left (16 \, b^{3} d^{3} x^{3} - b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} - a^{3} d^{3} + 24 \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} + 6 \,{\left (b^{3} c^{2} d + 6 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{3 \,{\left (a^{2} b^{4} c^{6} - 4 \, a^{3} b^{3} c^{5} d + 6 \, a^{4} b^{2} c^{4} d^{2} - 4 \, a^{5} b c^{3} d^{3} + a^{6} c^{2} d^{4} +{\left (b^{6} c^{4} d^{2} - 4 \, a b^{5} c^{3} d^{3} + 6 \, a^{2} b^{4} c^{2} d^{4} - 4 \, a^{3} b^{3} c d^{5} + a^{4} b^{2} d^{6}\right )} x^{4} + 2 \,{\left (b^{6} c^{5} d - 3 \, a b^{5} c^{4} d^{2} + 2 \, a^{2} b^{4} c^{3} d^{3} + 2 \, a^{3} b^{3} c^{2} d^{4} - 3 \, a^{4} b^{2} c d^{5} + a^{5} b d^{6}\right )} x^{3} +{\left (b^{6} c^{6} - 9 \, a^{2} b^{4} c^{4} d^{2} + 16 \, a^{3} b^{3} c^{3} d^{3} - 9 \, a^{4} b^{2} c^{2} d^{4} + a^{6} d^{6}\right )} x^{2} + 2 \,{\left (a b^{5} c^{6} - 3 \, a^{2} b^{4} c^{5} d + 2 \, a^{3} b^{3} c^{4} d^{2} + 2 \, a^{4} b^{2} c^{3} d^{3} - 3 \, a^{5} b c^{2} d^{4} + a^{6} c d^{5}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(16*b^3*d^3*x^3 - b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 - a^3*d^3 + 24*(b^3*c*d^2 + a*b^2*d^3)*x^2 + 6*(
b^3*c^2*d + 6*a*b^2*c*d^2 + a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b
^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*
d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*
d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 +
2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.77136, size = 718, normalized size = 5.32 \begin{align*} -\frac{\sqrt{b x + a}{\left (\frac{8 \,{\left (b^{7} c^{3} d^{4}{\left | b \right |} - 3 \, a b^{6} c^{2} d^{5}{\left | b \right |} + 3 \, a^{2} b^{5} c d^{6}{\left | b \right |} - a^{3} b^{4} d^{7}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}} + \frac{9 \,{\left (b^{8} c^{4} d^{3}{\left | b \right |} - 4 \, a b^{7} c^{3} d^{4}{\left | b \right |} + 6 \, a^{2} b^{6} c^{2} d^{5}{\left | b \right |} - 4 \, a^{3} b^{5} c d^{6}{\left | b \right |} + a^{4} b^{4} d^{7}{\left | b \right |}\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}}\right )}}{24 \,{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} + \frac{8 \,{\left (4 \, \sqrt{b d} b^{7} c^{2} d - 8 \, \sqrt{b d} a b^{6} c d^{2} + 4 \, \sqrt{b d} a^{2} b^{5} d^{3} - 9 \, \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 d + 9 \, \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} d^{2} + 3 \, \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} d\right )}}{3 \,{\left (b^{3} c^{3}{\left | b \right |} - 3 \, a b^{2} c^{2} d{\left | b \right |} + 3 \, a^{2} b c d^{2}{\left | b \right |} - a^{3} d^{3}{\left | b \right |}\right )}{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*sqrt(b*x + a)*(8*(b^7*c^3*d^4*abs(b) - 3*a*b^6*c^2*d^5*abs(b) + 3*a^2*b^5*c*d^6*abs(b) - a^3*b^4*d^7*abs
(b))*(b*x + a)/(b^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6) + 9*(b^8*c^4*d^3*abs(b) - 4*a*b^7*c^3*d^4*abs(b) +
6*a^2*b^6*c^2*d^5*abs(b) - 4*a^3*b^5*c*d^6*abs(b) + a^4*b^4*d^7*abs(b))/(b^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6
*d^6))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 8/3*(4*sqrt(b*d)*b^7*c^2*d - 8*sqrt(b*d)*a*b^6*c*d^2 + 4*sqrt(b
*d)*a^2*b^5*d^3 - 9*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*d + 9*sq
rt(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*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^3*d)/((b^3*c^3*abs(b) - 3*a*b^2*c^2*d*abs(b) + 3*a^2*b
*c*d^2*abs(b) - a^3*d^3*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)^3)