3.1518 \(\int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=98 \[ -\frac {16 b d \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^3}-\frac {8 d \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)^2}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 98, 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} \[ -\frac {16 b d \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^3}-\frac {8 d \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)^2}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (8*d*Sqrt[a + b*x])/(3*(b*c - a*d)^2*(c + d*x)^(3/2)) - (16*b
*d*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)^{3/2} (c+d x)^{5/2}} \, dx &=-\frac {2}{(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {(4 d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx}{b c-a d}\\ &=-\frac {2}{(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {8 d \sqrt {a+b x}}{3 (b c-a d)^2 (c+d x)^{3/2}}-\frac {(8 b d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 (b c-a d)^2}\\ &=-\frac {2}{(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {8 d \sqrt {a+b x}}{3 (b c-a d)^2 (c+d x)^{3/2}}-\frac {16 b d \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 = 78, normalized size = 0.80 \[ \frac {2 a^2 d^2-4 a b d (3 c+2 d x)-2 b^2 \left (3 c^2+12 c d x+8 d^2 x^2\right )}{3 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*d^2 - 4*a*b*d*(3*c + 2*d*x) - 2*b^2*(3*c^2 + 12*c*d*x + 8*d^2*x^2))/(3*(b*c - a*d)^3*Sqrt[a + b*x]*(c +
d*x)^(3/2))

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fricas [B]  time = 0.82, size = 273, normalized size = 2.79 \[ -\frac {2 \, {\left (8 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2} + 4 \, {\left (3 \, b^{2} c d + a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.48, size = 373, normalized size = 3.81 \[ -\frac {4 \, \sqrt {b d} b^{3}}{{\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {5 \, {\left (b^{6} c^{2} d^{3} {\left | b \right |} - 2 \, a b^{5} c d^{4} {\left | b \right |} + a^{2} b^{4} d^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{7} c^{5} d - 5 \, a b^{6} c^{4} d^{2} + 10 \, a^{2} b^{5} c^{3} d^{3} - 10 \, a^{3} b^{4} c^{2} d^{4} + 5 \, a^{4} b^{3} c d^{5} - a^{5} b^{2} d^{6}} + \frac {6 \, {\left (b^{7} c^{3} d^{2} {\left | b \right |} - 3 \, a b^{6} c^{2} d^{3} {\left | b \right |} + 3 \, a^{2} b^{5} c d^{4} {\left | b \right |} - a^{3} b^{4} d^{5} {\left | b \right |}\right )}}{b^{7} c^{5} d - 5 \, a b^{6} c^{4} d^{2} + 10 \, a^{2} b^{5} c^{3} d^{3} - 10 \, a^{3} b^{4} c^{2} d^{4} + 5 \, a^{4} b^{3} c d^{5} - a^{5} b^{2} d^{6}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 104, normalized size = 1.06 \[ -\frac {2 \left (-8 b^{2} x^{2} d^{2}-4 a b \,d^{2} x -12 b^{2} c d x +a^{2} d^{2}-6 a b c d -3 b^{2} c^{2}\right )}{3 \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-8*b^2*d^2*x^2-4*a*b*d^2*x-12*b^2*c*d*x+a^2*d^2-6*a*b*c*d-3*b^2*c^2)/(b*x+a)^(1/2)/(d*x+c)^(3/2)/(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(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 = 1.03, size = 132, normalized size = 1.35 \[ \frac {\sqrt {c+d\,x}\,\left (\frac {16\,b^2\,x^2}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {-2\,a^2\,d^2+12\,a\,b\,c\,d+6\,b^2\,c^2}{3\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,b\,x\,\left (a\,d+3\,b\,c\right )}{3\,d\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {c^2\,\sqrt {a+b\,x}}{d^2}+\frac {2\,c\,x\,\sqrt {a+b\,x}}{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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