3.1403 \(\int (a+b x) (c+d x)^{5/2} \, dx\)

Optimal. Leaf size=42 \[ \frac{2 b (c+d x)^{9/2}}{9 d^2}-\frac{2 (c+d x)^{7/2} (b c-a d)}{7 d^2} \]

[Out]

(-2*(b*c - a*d)*(c + d*x)^(7/2))/(7*d^2) + (2*b*(c + d*x)^(9/2))/(9*d^2)

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Rubi [A]  time = 0.0148131, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{2 b (c+d x)^{9/2}}{9 d^2}-\frac{2 (c+d x)^{7/2} (b c-a d)}{7 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*(c + d*x)^(5/2),x]

[Out]

(-2*(b*c - a*d)*(c + d*x)^(7/2))/(7*d^2) + (2*b*(c + d*x)^(9/2))/(9*d^2)

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 (a+b x) (c+d x)^{5/2} \, dx &=\int \left (\frac{(-b c+a d) (c+d x)^{5/2}}{d}+\frac{b (c+d x)^{7/2}}{d}\right ) \, dx\\ &=-\frac{2 (b c-a d) (c+d x)^{7/2}}{7 d^2}+\frac{2 b (c+d x)^{9/2}}{9 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0224357, size = 30, normalized size = 0.71 \[ \frac{2 (c+d x)^{7/2} (9 a d-2 b c+7 b d x)}{63 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c + d*x)^(7/2)*(-2*b*c + 9*a*d + 7*b*d*x))/(63*d^2)

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Maple [A]  time = 0.002, size = 27, normalized size = 0.6 \begin{align*}{\frac{14\,bdx+18\,ad-4\,bc}{63\,{d}^{2}} \left ( dx+c \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(d*x+c)^(7/2)*(7*b*d*x+9*a*d-2*b*c)/d^2

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Maxima [A]  time = 0.949677, size = 45, normalized size = 1.07 \begin{align*} \frac{2 \,{\left (7 \,{\left (d x + c\right )}^{\frac{9}{2}} b - 9 \,{\left (b c - a d\right )}{\left (d x + c\right )}^{\frac{7}{2}}\right )}}{63 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*(d*x + c)^(9/2)*b - 9*(b*c - a*d)*(d*x + c)^(7/2))/d^2

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Fricas [B]  time = 1.75748, size = 205, normalized size = 4.88 \begin{align*} \frac{2 \,{\left (7 \, b d^{4} x^{4} - 2 \, b c^{4} + 9 \, a c^{3} d +{\left (19 \, b c d^{3} + 9 \, a d^{4}\right )} x^{3} + 3 \,{\left (5 \, b c^{2} d^{2} + 9 \, a c d^{3}\right )} x^{2} +{\left (b c^{3} d + 27 \, a c^{2} d^{2}\right )} x\right )} \sqrt{d x + c}}{63 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*b*d^4*x^4 - 2*b*c^4 + 9*a*c^3*d + (19*b*c*d^3 + 9*a*d^4)*x^3 + 3*(5*b*c^2*d^2 + 9*a*c*d^3)*x^2 + (b*c^
3*d + 27*a*c^2*d^2)*x)*sqrt(d*x + c)/d^2

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Sympy [A]  time = 2.17768, size = 194, normalized size = 4.62 \begin{align*} \begin{cases} \frac{2 a c^{3} \sqrt{c + d x}}{7 d} + \frac{6 a c^{2} x \sqrt{c + d x}}{7} + \frac{6 a c d x^{2} \sqrt{c + d x}}{7} + \frac{2 a d^{2} x^{3} \sqrt{c + d x}}{7} - \frac{4 b c^{4} \sqrt{c + d x}}{63 d^{2}} + \frac{2 b c^{3} x \sqrt{c + d x}}{63 d} + \frac{10 b c^{2} x^{2} \sqrt{c + d x}}{21} + \frac{38 b c d x^{3} \sqrt{c + d x}}{63} + \frac{2 b d^{2} x^{4} \sqrt{c + d x}}{9} & \text{for}\: d \neq 0 \\c^{\frac{5}{2}} \left (a x + \frac{b x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a*c**3*sqrt(c + d*x)/(7*d) + 6*a*c**2*x*sqrt(c + d*x)/7 + 6*a*c*d*x**2*sqrt(c + d*x)/7 + 2*a*d**2
*x**3*sqrt(c + d*x)/7 - 4*b*c**4*sqrt(c + d*x)/(63*d**2) + 2*b*c**3*x*sqrt(c + d*x)/(63*d) + 10*b*c**2*x**2*sq
rt(c + d*x)/21 + 38*b*c*d*x**3*sqrt(c + d*x)/63 + 2*b*d**2*x**4*sqrt(c + d*x)/9, Ne(d, 0)), (c**(5/2)*(a*x + b
*x**2/2), True))

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Giac [B]  time = 1.07119, size = 263, normalized size = 6.26 \begin{align*} \frac{2 \,{\left (105 \,{\left (d x + c\right )}^{\frac{3}{2}} a c^{2} + 42 \,{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x + c\right )}^{\frac{3}{2}} c\right )} a c + \frac{21 \,{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x + c\right )}^{\frac{3}{2}} c\right )} b c^{2}}{d} + 3 \,{\left (15 \,{\left (d x + c\right )}^{\frac{7}{2}} - 42 \,{\left (d x + c\right )}^{\frac{5}{2}} c + 35 \,{\left (d x + c\right )}^{\frac{3}{2}} c^{2}\right )} a + \frac{6 \,{\left (15 \,{\left (d x + c\right )}^{\frac{7}{2}} - 42 \,{\left (d x + c\right )}^{\frac{5}{2}} c + 35 \,{\left (d x + c\right )}^{\frac{3}{2}} c^{2}\right )} b c}{d} + \frac{{\left (35 \,{\left (d x + c\right )}^{\frac{9}{2}} - 135 \,{\left (d x + c\right )}^{\frac{7}{2}} c + 189 \,{\left (d x + c\right )}^{\frac{5}{2}} c^{2} - 105 \,{\left (d x + c\right )}^{\frac{3}{2}} c^{3}\right )} b}{d}\right )}}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(105*(d*x + c)^(3/2)*a*c^2 + 42*(3*(d*x + c)^(5/2) - 5*(d*x + c)^(3/2)*c)*a*c + 21*(3*(d*x + c)^(5/2) -
5*(d*x + c)^(3/2)*c)*b*c^2/d + 3*(15*(d*x + c)^(7/2) - 42*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2)*a + 6*(1
5*(d*x + c)^(7/2) - 42*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2)*b*c/d + (35*(d*x + c)^(9/2) - 135*(d*x + c)
^(7/2)*c + 189*(d*x + c)^(5/2)*c^2 - 105*(d*x + c)^(3/2)*c^3)*b/d)/d