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

Optimal. Leaf size=71 \[ -\frac {4 b (c+d x)^{7/2} (b c-a d)}{7 d^3}+\frac {2 (c+d x)^{5/2} (b c-a d)^2}{5 d^3}+\frac {2 b^2 (c+d x)^{9/2}}{9 d^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 71, 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} \begin {gather*} -\frac {4 b (c+d x)^{7/2} (b c-a d)}{7 d^3}+\frac {2 (c+d x)^{5/2} (b c-a d)^2}{5 d^3}+\frac {2 b^2 (c+d x)^{9/2}}{9 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2*(c + d*x)^(3/2),x]

[Out]

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

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)^2 (c+d x)^{3/2} \, dx &=\int \left (\frac {(-b c+a d)^2 (c+d x)^{3/2}}{d^2}-\frac {2 b (b c-a d) (c+d x)^{5/2}}{d^2}+\frac {b^2 (c+d x)^{7/2}}{d^2}\right ) \, dx\\ &=\frac {2 (b c-a d)^2 (c+d x)^{5/2}}{5 d^3}-\frac {4 b (b c-a d) (c+d x)^{7/2}}{7 d^3}+\frac {2 b^2 (c+d x)^{9/2}}{9 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.86 \begin {gather*} \frac {2 (c+d x)^{5/2} \left (63 a^2 d^2+18 a b d (5 d x-2 c)+b^2 \left (8 c^2-20 c d x+35 d^2 x^2\right )\right )}{315 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c + d*x)^(5/2)*(63*a^2*d^2 + 18*a*b*d*(-2*c + 5*d*x) + b^2*(8*c^2 - 20*c*d*x + 35*d^2*x^2)))/(315*d^3)

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IntegrateAlgebraic [A]  time = 0.04, size = 72, normalized size = 1.01 \begin {gather*} \frac {2 (c+d x)^{5/2} \left (63 a^2 d^2+90 a b d (c+d x)-126 a b c d+63 b^2 c^2+35 b^2 (c+d x)^2-90 b^2 c (c+d x)\right )}{315 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c + d*x)^(5/2)*(63*b^2*c^2 - 126*a*b*c*d + 63*a^2*d^2 - 90*b^2*c*(c + d*x) + 90*a*b*d*(c + d*x) + 35*b^2*(
c + d*x)^2))/(315*d^3)

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fricas [B]  time = 1.24, size = 137, normalized size = 1.93 \begin {gather*} \frac {2 \, {\left (35 \, b^{2} d^{4} x^{4} + 8 \, b^{2} c^{4} - 36 \, a b c^{3} d + 63 \, a^{2} c^{2} d^{2} + 10 \, {\left (5 \, b^{2} c d^{3} + 9 \, a b d^{4}\right )} x^{3} + 3 \, {\left (b^{2} c^{2} d^{2} + 48 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{2} - 2 \, {\left (2 \, b^{2} c^{3} d - 9 \, a b c^{2} d^{2} - 63 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^2*d^4*x^4 + 8*b^2*c^4 - 36*a*b*c^3*d + 63*a^2*c^2*d^2 + 10*(5*b^2*c*d^3 + 9*a*b*d^4)*x^3 + 3*(b^2*
c^2*d^2 + 48*a*b*c*d^3 + 21*a^2*d^4)*x^2 - 2*(2*b^2*c^3*d - 9*a*b*c^2*d^2 - 63*a^2*c*d^3)*x)*sqrt(d*x + c)/d^3

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giac [B]  time = 1.41, size = 360, normalized size = 5.07 \begin {gather*} \frac {2 \, {\left (315 \, \sqrt {d x + c} a^{2} c^{2} + 210 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{2} c + \frac {210 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b c^{2}}{d} + 21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a^{2} + \frac {21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b^{2} c^{2}}{d^{2}} + \frac {84 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b c}{d} + \frac {18 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b^{2} c}{d^{2}} + \frac {18 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} a b}{d} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b^{2}}{d^{2}}\right )}}{315 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(315*sqrt(d*x + c)*a^2*c^2 + 210*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^2*c + 210*((d*x + c)^(3/2) - 3*
sqrt(d*x + c)*c)*a*b*c^2/d + 21*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^2 + 21*(3*
(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*b^2*c^2/d^2 + 84*(3*(d*x + c)^(5/2) - 10*(d*x +
c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a*b*c/d + 18*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2
)*c^2 - 35*sqrt(d*x + c)*c^3)*b^2*c/d^2 + 18*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^
2 - 35*sqrt(d*x + c)*c^3)*a*b/d + (35*(d*x + c)^(9/2) - 180*(d*x + c)^(7/2)*c + 378*(d*x + c)^(5/2)*c^2 - 420*
(d*x + c)^(3/2)*c^3 + 315*sqrt(d*x + c)*c^4)*b^2/d^2)/d

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maple [A]  time = 0.01, size = 63, normalized size = 0.89 \begin {gather*} \frac {2 \left (d x +c \right )^{\frac {5}{2}} \left (35 b^{2} x^{2} d^{2}+90 a b \,d^{2} x -20 b^{2} c d x +63 a^{2} d^{2}-36 a b c d +8 b^{2} c^{2}\right )}{315 d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(d*x+c)^(5/2)*(35*b^2*d^2*x^2+90*a*b*d^2*x-20*b^2*c*d*x+63*a^2*d^2-36*a*b*c*d+8*b^2*c^2)/d^3

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maxima [A]  time = 1.38, size = 68, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{2} - 90 \, {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}^{\frac {7}{2}} + 63 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}}\right )}}{315 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(d*x + c)^(9/2)*b^2 - 90*(b^2*c - a*b*d)*(d*x + c)^(7/2) + 63*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x +
c)^(5/2))/d^3

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mupad [B]  time = 0.06, size = 68, normalized size = 0.96 \begin {gather*} \frac {2\,{\left (c+d\,x\right )}^{5/2}\,\left (35\,b^2\,{\left (c+d\,x\right )}^2+63\,a^2\,d^2+63\,b^2\,c^2-90\,b^2\,c\,\left (c+d\,x\right )+90\,a\,b\,d\,\left (c+d\,x\right )-126\,a\,b\,c\,d\right )}{315\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2*(c + d*x)^(3/2),x)

[Out]

(2*(c + d*x)^(5/2)*(35*b^2*(c + d*x)^2 + 63*a^2*d^2 + 63*b^2*c^2 - 90*b^2*c*(c + d*x) + 90*a*b*d*(c + d*x) - 1
26*a*b*c*d))/(315*d^3)

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sympy [A]  time = 9.61, size = 240, normalized size = 3.38 \begin {gather*} a^{2} c \left (\begin {cases} \sqrt {c} x & \text {for}\: d = 0 \\\frac {2 \left (c + d x\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + \frac {2 a^{2} \left (- \frac {c \left (c + d x\right )^{\frac {3}{2}}}{3} + \frac {\left (c + d x\right )^{\frac {5}{2}}}{5}\right )}{d} + \frac {4 a b c \left (- \frac {c \left (c + d x\right )^{\frac {3}{2}}}{3} + \frac {\left (c + d x\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} + \frac {4 a b \left (\frac {c^{2} \left (c + d x\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + d x\right )^{\frac {5}{2}}}{5} + \frac {\left (c + d x\right )^{\frac {7}{2}}}{7}\right )}{d^{2}} + \frac {2 b^{2} c \left (\frac {c^{2} \left (c + d x\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + d x\right )^{\frac {5}{2}}}{5} + \frac {\left (c + d x\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} + \frac {2 b^{2} \left (- \frac {c^{3} \left (c + d x\right )^{\frac {3}{2}}}{3} + \frac {3 c^{2} \left (c + d x\right )^{\frac {5}{2}}}{5} - \frac {3 c \left (c + d x\right )^{\frac {7}{2}}}{7} + \frac {\left (c + d x\right )^{\frac {9}{2}}}{9}\right )}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c*Piecewise((sqrt(c)*x, Eq(d, 0)), (2*(c + d*x)**(3/2)/(3*d), True)) + 2*a**2*(-c*(c + d*x)**(3/2)/3 + (c
+ d*x)**(5/2)/5)/d + 4*a*b*c*(-c*(c + d*x)**(3/2)/3 + (c + d*x)**(5/2)/5)/d**2 + 4*a*b*(c**2*(c + d*x)**(3/2)
/3 - 2*c*(c + d*x)**(5/2)/5 + (c + d*x)**(7/2)/7)/d**2 + 2*b**2*c*(c**2*(c + d*x)**(3/2)/3 - 2*c*(c + d*x)**(5
/2)/5 + (c + d*x)**(7/2)/7)/d**3 + 2*b**2*(-c**3*(c + d*x)**(3/2)/3 + 3*c**2*(c + d*x)**(5/2)/5 - 3*c*(c + d*x
)**(7/2)/7 + (c + d*x)**(9/2)/9)/d**3

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