3.1436 \(\int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ -\frac {6 b^2 \sqrt {c+d x} (b c-a d)}{d^4}-\frac {6 b (b c-a d)^2}{d^4 \sqrt {c+d x}}+\frac {2 (b c-a d)^3}{3 d^4 (c+d x)^{3/2}}+\frac {2 b^3 (c+d x)^{3/2}}{3 d^4} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {43} \[ -\frac {6 b^2 \sqrt {c+d x} (b c-a d)}{d^4}-\frac {6 b (b c-a d)^2}{d^4 \sqrt {c+d x}}+\frac {2 (b c-a d)^3}{3 d^4 (c+d x)^{3/2}}+\frac {2 b^3 (c+d x)^{3/2}}{3 d^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/(c + d*x)^(5/2),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx &=\int \left (\frac {(-b c+a d)^3}{d^3 (c+d x)^{5/2}}+\frac {3 b (b c-a d)^2}{d^3 (c+d x)^{3/2}}-\frac {3 b^2 (b c-a d)}{d^3 \sqrt {c+d x}}+\frac {b^3 \sqrt {c+d x}}{d^3}\right ) \, dx\\ &=\frac {2 (b c-a d)^3}{3 d^4 (c+d x)^{3/2}}-\frac {6 b (b c-a d)^2}{d^4 \sqrt {c+d x}}-\frac {6 b^2 (b c-a d) \sqrt {c+d x}}{d^4}+\frac {2 b^3 (c+d x)^{3/2}}{3 d^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 76, normalized size = 0.79 \[ \frac {2 \left (-9 b^2 (c+d x)^2 (b c-a d)-9 b (c+d x) (b c-a d)^2+(b c-a d)^3+b^3 (c+d x)^3\right )}{3 d^4 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 136, normalized size = 1.42 \[ \frac {2 \, {\left (b^{3} d^{3} x^{3} - 16 \, b^{3} c^{3} + 24 \, a b^{2} c^{2} d - 6 \, a^{2} b c d^{2} - a^{3} d^{3} - 3 \, {\left (2 \, b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} c^{2} d - 12 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b^3*d^3*x^3 - 16*b^3*c^3 + 24*a*b^2*c^2*d - 6*a^2*b*c*d^2 - a^3*d^3 - 3*(2*b^3*c*d^2 - 3*a*b^2*d^3)*x^2 -
3*(8*b^3*c^2*d - 12*a*b^2*c*d^2 + 3*a^2*b*d^3)*x)*sqrt(d*x + c)/(d^6*x^2 + 2*c*d^5*x + c^2*d^4)

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giac [A]  time = 1.00, size = 141, normalized size = 1.47 \[ -\frac {2 \, {\left (9 \, {\left (d x + c\right )} b^{3} c^{2} - b^{3} c^{3} - 18 \, {\left (d x + c\right )} a b^{2} c d + 3 \, a b^{2} c^{2} d + 9 \, {\left (d x + c\right )} a^{2} b d^{2} - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{3} d^{8} - 9 \, \sqrt {d x + c} b^{3} c d^{8} + 9 \, \sqrt {d x + c} a b^{2} d^{9}\right )}}{3 \, d^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(9*(d*x + c)*b^3*c^2 - b^3*c^3 - 18*(d*x + c)*a*b^2*c*d + 3*a*b^2*c^2*d + 9*(d*x + c)*a^2*b*d^2 - 3*a^2*b
*c*d^2 + a^3*d^3)/((d*x + c)^(3/2)*d^4) + 2/3*((d*x + c)^(3/2)*b^3*d^8 - 9*sqrt(d*x + c)*b^3*c*d^8 + 9*sqrt(d*
x + c)*a*b^2*d^9)/d^12

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maple [A]  time = 0.01, size = 115, normalized size = 1.20 \[ -\frac {2 \left (-b^{3} x^{3} d^{3}-9 a \,b^{2} d^{3} x^{2}+6 b^{3} c \,d^{2} x^{2}+9 a^{2} b \,d^{3} x -36 a \,b^{2} c \,d^{2} x +24 b^{3} c^{2} d x +a^{3} d^{3}+6 a^{2} b c \,d^{2}-24 a \,b^{2} c^{2} d +16 b^{3} c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 122, normalized size = 1.27 \[ \frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} b^{3} - 9 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {d x + c}}{d^{3}} + \frac {b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3} - 9 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (d x + c\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{3}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 128, normalized size = 1.33 \[ \frac {2\,b^3\,{\left (c+d\,x\right )}^3-2\,a^3\,d^3+2\,b^3\,c^3-18\,b^3\,c\,{\left (c+d\,x\right )}^2-18\,b^3\,c^2\,\left (c+d\,x\right )+18\,a\,b^2\,d\,{\left (c+d\,x\right )}^2-18\,a^2\,b\,d^2\,\left (c+d\,x\right )-6\,a\,b^2\,c^2\,d+6\,a^2\,b\,c\,d^2+36\,a\,b^2\,c\,d\,\left (c+d\,x\right )}{3\,d^4\,{\left (c+d\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^3/(c + d*x)^(5/2),x)

[Out]

(2*b^3*(c + d*x)^3 - 2*a^3*d^3 + 2*b^3*c^3 - 18*b^3*c*(c + d*x)^2 - 18*b^3*c^2*(c + d*x) + 18*a*b^2*d*(c + d*x
)^2 - 18*a^2*b*d^2*(c + d*x) - 6*a*b^2*c^2*d + 6*a^2*b*c*d^2 + 36*a*b^2*c*d*(c + d*x))/(3*d^4*(c + d*x)^(3/2))

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sympy [A]  time = 1.44, size = 461, normalized size = 4.80 \[ \begin {cases} - \frac {2 a^{3} d^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {12 a^{2} b c d^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {18 a^{2} b d^{3} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {48 a b^{2} c^{2} d}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {72 a b^{2} c d^{2} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {18 a b^{2} d^{3} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {32 b^{3} c^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {48 b^{3} c^{2} d x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {12 b^{3} c d^{2} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {2 b^{3} d^{3} x^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} & \text {for}\: d \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a**3*d**3/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 12*a**2*b*c*d**2/(3*c*d**4*sqrt(c
+ d*x) + 3*d**5*x*sqrt(c + d*x)) - 18*a**2*b*d**3*x/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 48*a*b
**2*c**2*d/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 72*a*b**2*c*d**2*x/(3*c*d**4*sqrt(c + d*x) + 3*
d**5*x*sqrt(c + d*x)) + 18*a*b**2*d**3*x**2/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 32*b**3*c**3/(
3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 48*b**3*c**2*d*x/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c
+ d*x)) - 12*b**3*c*d**2*x**2/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 2*b**3*d**3*x**3/(3*c*d**4*s
qrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)), Ne(d, 0)), ((a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4/4)/c**
(5/2), True))

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