3.15.2 \(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=101 \[ \frac {16 d^2 \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^3}+\frac {8 d}{3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.02, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {16 d^2 \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^3}+\frac {8 d}{3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}-\frac {2}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.74 \begin {gather*} \frac {2 \left (3 a^2 d^2+6 a b d (c+2 d x)+b^2 \left (-c^2+4 c d x+8 d^2 x^2\right )\right )}{3 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x]

[Out]

(2*(3*a^2*d^2 + 6*a*b*d*(c + 2*d*x) + b^2*(-c^2 + 4*c*d*x + 8*d^2*x^2)))/(3*(b*c - a*d)^3*(a + b*x)^(3/2)*Sqrt
[c + d*x])

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IntegrateAlgebraic [A]  time = 0.12, size = 73, normalized size = 0.72 \begin {gather*} \frac {2 (c+d x)^{3/2} \left (\frac {3 d^2 (a+b x)^2}{(c+d x)^2}+\frac {6 b d (a+b x)}{c+d x}-b^2\right )}{3 (a+b x)^{3/2} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x]

[Out]

(2*(c + d*x)^(3/2)*(-b^2 + (3*d^2*(a + b*x)^2)/(c + d*x)^2 + (6*b*d*(a + b*x))/(c + d*x)))/(3*(b*c - a*d)^3*(a
+ b*x)^(3/2))

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fricas [B]  time = 1.92, size = 273, normalized size = 2.70 \begin {gather*} \frac {2 \, {\left (8 \, b^{2} d^{2} x^{2} - b^{2} c^{2} + 6 \, a b c d + 3 \, a^{2} d^{2} + 4 \, {\left (b^{2} c d + 3 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.54, size = 368, normalized size = 3.64 \begin {gather*} \frac {2 \, \sqrt {b x + a} b^{2} d^{2}}{{\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 )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {4 \, {\left (5 \, \sqrt {b d} b^{6} c^{2} d - 10 \, \sqrt {b d} a b^{5} c d^{2} + 5 \, \sqrt {b d} a^{2} b^{4} d^{3} - 12 \, \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^{4} c d + 12 \, \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^{3} 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^{2} d\right )}}{3 \, {\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\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)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(b*x + a)*b^2*d^2/((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))*sqrt(
b^2*c + (b*x + a)*b*d - a*b*d)) + 4/3*(5*sqrt(b*d)*b^6*c^2*d - 10*sqrt(b*d)*a*b^5*c*d^2 + 5*sqrt(b*d)*a^2*b^4*
d^3 - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^4*c*d + 12*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^2*d)/((b^2*c^2*abs(b) - 2*a*b*c*d*abs(b) + a^2*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)^3)

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \begin {gather*} -\frac {2 \left (8 b^{2} x^{2} d^{2}+12 a b \,d^{2} x +4 b^{2} c d x +3 a^{2} d^{2}+6 a b c d -b^{2} c^{2}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}\, \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(5/2)/(d*x+c)^(3/2),x)

[Out]

-2/3*(8*b^2*d^2*x^2+12*a*b*d^2*x+4*b^2*c*d*x+3*a^2*d^2+6*a*b*c*d-b^2*c^2)/(b*x+a)^(3/2)/(d*x+c)^(1/2)/(a^3*d^3
-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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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)^(3/2),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 = 1.06, size = 141, normalized size = 1.40 \begin {gather*} -\frac {\sqrt {c+d\,x}\,\left (\frac {8\,x\,\left (3\,a\,d+b\,c\right )}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {16\,b\,d\,x^2}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {6\,a^2\,d^2+12\,a\,b\,c\,d-2\,b^2\,c^2}{3\,b\,d\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {a\,c\,\sqrt {a+b\,x}}{b\,d}+\frac {x\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(5/2)*(c + d*x)^(3/2)),x)

[Out]

-((c + d*x)^(1/2)*((8*x*(3*a*d + b*c))/(3*(a*d - b*c)^3) + (16*b*d*x^2)/(3*(a*d - b*c)^3) + (6*a^2*d^2 - 2*b^2
*c^2 + 12*a*b*c*d)/(3*b*d*(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/2) + (a*c*(a + b*x)^(1/2))/(b*d) + (x*(a*d + b*c)
*(a + b*x)^(1/2))/(b*d))

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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 {3}{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)**(3/2),x)

[Out]

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(3/2)), x)

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