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

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

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Rubi [A]  time = 0.03, antiderivative size = 135, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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} (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*}

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Mathematica [A]  time = 0.04, size = 118, normalized size = 0.87 \begin {gather*} \frac {-2 a^3 d^3+6 a^2 b d^2 (3 c+2 d x)+6 a b^2 d \left (3 c^2+12 c d x+8 d^2 x^2\right )+b^3 \left (-2 c^3+12 c^2 d x+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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.00, size = 92, normalized size = 0.68 \begin {gather*} -\frac {2 (a+b x)^{3/2} \left (\frac {b^3 (c+d x)^3}{(a+b x)^3}-\frac {9 b^2 d (c+d x)^2}{(a+b x)^2}-\frac {9 b d^2 (c+d x)}{a+b x}+d^3\right )}{3 (c+d x)^{3/2} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 4.16, size = 447, normalized size = 3.31 \begin {gather*} \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 {gather*}

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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giac [B]  time = 2.66, size = 670, normalized size = 4.96 \begin {gather*} \frac {2 \, \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^{9} c^{7} d - 7 \, a b^{8} c^{6} d^{2} + 21 \, a^{2} b^{7} c^{5} d^{3} - 35 \, a^{3} b^{6} c^{4} d^{4} + 35 \, a^{4} b^{5} c^{3} d^{5} - 21 \, a^{5} b^{4} c^{2} d^{6} + 7 \, a^{6} b^{3} c d^{7} - a^{7} b^{2} d^{8}} + \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^{9} c^{7} d - 7 \, a b^{8} c^{6} d^{2} + 21 \, a^{2} b^{7} c^{5} d^{3} - 35 \, a^{3} b^{6} c^{4} d^{4} + 35 \, a^{4} b^{5} c^{3} d^{5} - 21 \, a^{5} b^{4} c^{2} d^{6} + 7 \, a^{6} b^{3} c d^{7} - a^{7} b^{2} d^{8}}\right )}}{3 \, {\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 {gather*}

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]

2/3*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^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*
a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8) + 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^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^
3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8))/(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*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*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(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)

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maple [A]  time = 0.01, size = 169, normalized size = 1.25 \begin {gather*} -\frac {2 \left (-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}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \left (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}\right )} \end {gather*}

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 = 0.00, 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)^(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 = 1.85, size = 224, normalized size = 1.66 \begin {gather*} \frac {\sqrt {c+d\,x}\,\left (\frac {16\,b\,x^2\,\left (a\,d+b\,c\right )}{{\left (a\,d-b\,c\right )}^4}-\frac {2\,a^3\,d^3-18\,a^2\,b\,c\,d^2-18\,a\,b^2\,c^2\,d+2\,b^3\,c^3}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}+\frac {32\,b^2\,d\,x^3}{3\,{\left (a\,d-b\,c\right )}^4}+\frac {4\,x\,\left (a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{d\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a\,c^2\,\sqrt {a+b\,x}}{b\,d^2}+\frac {x^2\,\left (a\,d+2\,b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {c\,x\,\left (2\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*x)^(1/2)*((16*b*x^2*(a*d + b*c))/(a*d - b*c)^4 - (2*a^3*d^3 + 2*b^3*c^3 - 18*a*b^2*c^2*d - 18*a^2*b*c*
d^2)/(3*b*d^2*(a*d - b*c)^4) + (32*b^2*d*x^3)/(3*(a*d - b*c)^4) + (4*x*(a^2*d^2 + b^2*c^2 + 6*a*b*c*d))/(d*(a*
d - b*c)^4)))/(x^3*(a + b*x)^(1/2) + (a*c^2*(a + b*x)^(1/2))/(b*d^2) + (x^2*(a*d + 2*b*c)*(a + b*x)^(1/2))/(b*
d) + (c*x*(2*a*d + b*c)*(a + b*x)^(1/2))/(b*d^2))

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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}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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