3.4.77 \(\int (a+b x)^{3/2} (A+B x) \, dx\)

Optimal. Leaf size=42 \[ \frac {2 (a+b x)^{5/2} (A b-a B)}{5 b^2}+\frac {2 B (a+b x)^{7/2}}{7 b^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {2 (a+b x)^{5/2} (A b-a B)}{5 b^2}+\frac {2 B (a+b x)^{7/2}}{7 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)*(A + B*x),x]

[Out]

(2*(A*b - a*B)*(a + b*x)^(5/2))/(5*b^2) + (2*B*(a + b*x)^(7/2))/(7*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)^{3/2} (A+B x) \, dx &=\int \left (\frac {(A b-a B) (a+b x)^{3/2}}{b}+\frac {B (a+b x)^{5/2}}{b}\right ) \, dx\\ &=\frac {2 (A b-a B) (a+b x)^{5/2}}{5 b^2}+\frac {2 B (a+b x)^{7/2}}{7 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 0.71 \begin {gather*} \frac {2 (a+b x)^{5/2} (-2 a B+7 A b+5 b B x)}{35 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)*(A + B*x),x]

[Out]

(2*(a + b*x)^(5/2)*(7*A*b - 2*a*B + 5*b*B*x))/(35*b^2)

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IntegrateAlgebraic [A]  time = 0.03, size = 33, normalized size = 0.79 \begin {gather*} \frac {2 (a+b x)^{5/2} (5 B (a+b x)-7 a B+7 A b)}{35 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(3/2)*(A + B*x),x]

[Out]

(2*(a + b*x)^(5/2)*(7*A*b - 7*a*B + 5*B*(a + b*x)))/(35*b^2)

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fricas [B]  time = 1.22, size = 69, normalized size = 1.64 \begin {gather*} \frac {2 \, {\left (5 \, B b^{3} x^{3} - 2 \, B a^{3} + 7 \, A a^{2} b + {\left (8 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2} + {\left (B a^{2} b + 14 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}}{35 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A),x, algorithm="fricas")

[Out]

2/35*(5*B*b^3*x^3 - 2*B*a^3 + 7*A*a^2*b + (8*B*a*b^2 + 7*A*b^3)*x^2 + (B*a^2*b + 14*A*a*b^2)*x)*sqrt(b*x + a)/
b^2

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giac [B]  time = 1.30, size = 192, normalized size = 4.57 \begin {gather*} \frac {2 \, {\left (105 \, \sqrt {b x + a} A a^{2} + 70 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A a + \frac {35 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} B a^{2}}{b} + 7 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A + \frac {14 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B a}{b} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B}{b}\right )}}{105 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A),x, algorithm="giac")

[Out]

2/105*(105*sqrt(b*x + a)*A*a^2 + 70*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*A*a + 35*((b*x + a)^(3/2) - 3*sqrt(b
*x + a)*a)*B*a^2/b + 7*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*A + 14*(3*(b*x + a)^(
5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*B*a/b + 3*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(
b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*B/b)/b

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maple [A]  time = 0.00, size = 27, normalized size = 0.64 \begin {gather*} \frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (5 B b x +7 A b -2 B a \right )}{35 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A),x)

[Out]

2/35*(b*x+a)^(5/2)*(5*B*b*x+7*A*b-2*B*a)/b^2

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maxima [A]  time = 0.79, size = 33, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} B - 7 \, {\left (B a - A b\right )} {\left (b x + a\right )}^{\frac {5}{2}}\right )}}{35 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A),x, algorithm="maxima")

[Out]

2/35*(5*(b*x + a)^(7/2)*B - 7*(B*a - A*b)*(b*x + a)^(5/2))/b^2

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mupad [B]  time = 0.04, size = 29, normalized size = 0.69 \begin {gather*} \frac {2\,{\left (a+b\,x\right )}^{5/2}\,\left (7\,A\,b-7\,B\,a+5\,B\,\left (a+b\,x\right )\right )}{35\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)*(a + b*x)^(3/2),x)

[Out]

(2*(a + b*x)^(5/2)*(7*A*b - 7*B*a + 5*B*(a + b*x)))/(35*b^2)

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sympy [A]  time = 0.69, size = 146, normalized size = 3.48 \begin {gather*} \begin {cases} \frac {2 A a^{2} \sqrt {a + b x}}{5 b} + \frac {4 A a x \sqrt {a + b x}}{5} + \frac {2 A b x^{2} \sqrt {a + b x}}{5} - \frac {4 B a^{3} \sqrt {a + b x}}{35 b^{2}} + \frac {2 B a^{2} x \sqrt {a + b x}}{35 b} + \frac {16 B a x^{2} \sqrt {a + b x}}{35} + \frac {2 B b x^{3} \sqrt {a + b x}}{7} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{2}}{2}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A),x)

[Out]

Piecewise((2*A*a**2*sqrt(a + b*x)/(5*b) + 4*A*a*x*sqrt(a + b*x)/5 + 2*A*b*x**2*sqrt(a + b*x)/5 - 4*B*a**3*sqrt
(a + b*x)/(35*b**2) + 2*B*a**2*x*sqrt(a + b*x)/(35*b) + 16*B*a*x**2*sqrt(a + b*x)/35 + 2*B*b*x**3*sqrt(a + b*x
)/7, Ne(b, 0)), (a**(3/2)*(A*x + B*x**2/2), True))

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