3.414 \(\int (a+b x)^{5/2} (A+B x) \, dx\)

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

[Out]

(2*(A*b - a*B)*(a + b*x)^(7/2))/(7*b^2) + (2*B*(a + b*x)^(9/2))/(9*b^2)

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Rubi [A]  time = 0.0141924, 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} \[ \frac{2 (a+b x)^{7/2} (A b-a B)}{7 b^2}+\frac{2 B (a+b x)^{9/2}}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)*(A + B*x),x]

[Out]

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

Mathematica [A]  time = 0.0215187, size = 30, normalized size = 0.71 \[ \frac{2 (a+b x)^{7/2} (-2 a B+9 A b+7 b B x)}{63 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)*(A + B*x),x]

[Out]

(2*(a + b*x)^(7/2)*(9*A*b - 2*a*B + 7*b*B*x))/(63*b^2)

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Maple [A]  time = 0.003, size = 27, normalized size = 0.6 \begin{align*}{\frac{14\,bBx+18\,Ab-4\,Ba}{63\,{b}^{2}} \left ( bx+a \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A),x)

[Out]

2/63*(b*x+a)^(7/2)*(7*B*b*x+9*A*b-2*B*a)/b^2

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Maxima [A]  time = 1.52742, size = 45, normalized size = 1.07 \begin{align*} \frac{2 \,{\left (7 \,{\left (b x + a\right )}^{\frac{9}{2}} B - 9 \,{\left (B a - A b\right )}{\left (b x + a\right )}^{\frac{7}{2}}\right )}}{63 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*(b*x + a)^(9/2)*B - 9*(B*a - A*b)*(b*x + a)^(7/2))/b^2

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Fricas [B]  time = 2.34175, size = 205, normalized size = 4.88 \begin{align*} \frac{2 \,{\left (7 \, B b^{4} x^{4} - 2 \, B a^{4} + 9 \, A a^{3} b +{\left (19 \, B a b^{3} + 9 \, A b^{4}\right )} x^{3} + 3 \,{\left (5 \, B a^{2} b^{2} + 9 \, A a b^{3}\right )} x^{2} +{\left (B a^{3} b + 27 \, A a^{2} b^{2}\right )} x\right )} \sqrt{b x + a}}{63 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.38924, size = 194, normalized size = 4.62 \begin{align*} \begin{cases} \frac{2 A a^{3} \sqrt{a + b x}}{7 b} + \frac{6 A a^{2} x \sqrt{a + b x}}{7} + \frac{6 A a b x^{2} \sqrt{a + b x}}{7} + \frac{2 A b^{2} x^{3} \sqrt{a + b x}}{7} - \frac{4 B a^{4} \sqrt{a + b x}}{63 b^{2}} + \frac{2 B a^{3} x \sqrt{a + b x}}{63 b} + \frac{10 B a^{2} x^{2} \sqrt{a + b x}}{21} + \frac{38 B a b x^{3} \sqrt{a + b x}}{63} + \frac{2 B b^{2} x^{4} \sqrt{a + b x}}{9} & \text{for}\: b \neq 0 \\a^{\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)**(5/2)*(B*x+A),x)

[Out]

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

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Giac [B]  time = 1.28132, size = 263, normalized size = 6.26 \begin{align*} \frac{2 \,{\left (105 \,{\left (b x + a\right )}^{\frac{3}{2}} A a^{2} + 42 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} A a + \frac{21 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} B a^{2}}{b} + 3 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} A + \frac{6 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} B a}{b} + \frac{{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} B}{b}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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