3.19.48 \(\int (a+b x)^m (c+d x) \, dx\) [1848]

Optimal. Leaf size=46 \[ \frac {(b c-a d) (a+b x)^{1+m}}{b^2 (1+m)}+\frac {d (a+b x)^{2+m}}{b^2 (2+m)} \]

[Out]

(-a*d+b*c)*(b*x+a)^(1+m)/b^2/(1+m)+d*(b*x+a)^(2+m)/b^2/(2+m)

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Rubi [A]
time = 0.01, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \begin {gather*} \frac {(b c-a d) (a+b x)^{m+1}}{b^2 (m+1)}+\frac {d (a+b x)^{m+2}}{b^2 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x),x]

[Out]

((b*c - a*d)*(a + b*x)^(1 + m))/(b^2*(1 + m)) + (d*(a + b*x)^(2 + m))/(b^2*(2 + m))

Rule 45

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)^m (c+d x) \, dx &=\int \left (\frac {(b c-a d) (a+b x)^m}{b}+\frac {d (a+b x)^{1+m}}{b}\right ) \, dx\\ &=\frac {(b c-a d) (a+b x)^{1+m}}{b^2 (1+m)}+\frac {d (a+b x)^{2+m}}{b^2 (2+m)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 41, normalized size = 0.89 \begin {gather*} \frac {(a+b x)^{1+m} (-a d+b c (2+m)+b d (1+m) x)}{b^2 (1+m) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x),x]

[Out]

((a + b*x)^(1 + m)*(-(a*d) + b*c*(2 + m) + b*d*(1 + m)*x))/(b^2*(1 + m)*(2 + m))

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.04, size = 383, normalized size = 8.33 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {x \left (2 c+d x\right ) a^m}{2},b\text {==}0\right \},\left \{\frac {a d \left (1+\text {Log}\left [\frac {a+b x}{b}\right ]\right )+b \left (-c+d x \text {Log}\left [\frac {a+b x}{b}\right ]\right )}{b^2 \left (a+b x\right )},m\text {==}-2\right \},\left \{\frac {-a d \text {Log}\left [\frac {a}{b}+x\right ]+b c \text {Log}\left [\frac {a}{b}+x\right ]+b d x}{b^2},m\text {==}-1\right \}\right \},-\frac {a^2 d \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {2 a b c \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {a b c m \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {a b d m x \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {2 b^2 c x \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {b^2 c m x \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {b^2 d x^2 \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}+\frac {b^2 d m x^2 \left (a+b x\right )^m}{2 b^2+3 b^2 m+b^2 m^2}\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(a + b*x)^m*(c + d*x)^1,x]')

[Out]

Piecewise[{{x (2 c + d x) a ^ m / 2, b == 0}, {(a d (1 + Log[(a + b x) / b]) + b (-c + d x Log[(a + b x) / b])
) / (b ^ 2 (a + b x)), m == -2}, {(-a d Log[a / b + x] + b c Log[a / b + x] + b d x) / b ^ 2, m == -1}}, -a ^
2 d (a + b x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2 m ^ 2) + 2 a b c (a + b x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2
m ^ 2) + a b c m (a + b x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2 m ^ 2) + a b d m x (a + b x) ^ m / (2 b ^ 2 + 3
b ^ 2 m + b ^ 2 m ^ 2) + 2 b ^ 2 c x (a + b x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2 m ^ 2) + b ^ 2 c m x (a + b
x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2 m ^ 2) + b ^ 2 d x ^ 2 (a + b x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2 m ^
2) + b ^ 2 d m x ^ 2 (a + b x) ^ m / (2 b ^ 2 + 3 b ^ 2 m + b ^ 2 m ^ 2)]

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Maple [A]
time = 0.15, size = 49, normalized size = 1.07

method	result	size
gosper	\(-\frac {\left (b x +a \right )^{1+m} \left (-b d m x -b c m -b d x +a d -2 b c \right )}{b^{2} \left (m^{2}+3 m +2\right )}\)	\(49\)
risch	\(-\frac {\left (-d \,x^{2} b^{2} m -a b d m x -b^{2} c m x -d \,x^{2} b^{2}-a b c m -2 b^{2} c x +a^{2} d -2 a b c \right ) \left (b x +a \right )^{m}}{b^{2} \left (2+m \right ) \left (1+m \right )}\)	\(81\)
norman	\(\frac {d \,x^{2} {\mathrm e}^{m \ln \left (b x +a \right )}}{2+m}+\frac {\left (a d m +b c m +2 b c \right ) x \,{\mathrm e}^{m \ln \left (b x +a \right )}}{b \left (m^{2}+3 m +2\right )}-\frac {a \left (-b c m +a d -2 b c \right ) {\mathrm e}^{m \ln \left (b x +a \right )}}{b^{2} \left (m^{2}+3 m +2\right )}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 63, normalized size = 1.37 \begin {gather*} \frac {{\left (b^{2} {\left (m + 1\right )} x^{2} + a b m x - a^{2}\right )} {\left (b x + a\right )}^{m} d}{{\left (m^{2} + 3 \, m + 2\right )} b^{2}} + \frac {{\left (b x + a\right )}^{m + 1} c}{b {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 83, normalized size = 1.80 \begin {gather*} \frac {{\left (a b c m + 2 \, a b c - a^{2} d + {\left (b^{2} d m + b^{2} d\right )} x^{2} + {\left (2 \, b^{2} c + {\left (b^{2} c + a b d\right )} m\right )} x\right )} {\left (b x + a\right )}^{m}}{b^{2} m^{2} + 3 \, b^{2} m + 2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.44, size = 377, normalized size = 8.20 \begin {gather*} \begin {cases} a^{m} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: b = 0 \\\frac {a d \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac {a d}{a b^{2} + b^{3} x} - \frac {b c}{a b^{2} + b^{3} x} + \frac {b d x \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text {for}\: m = -2 \\- \frac {a d \log {\left (\frac {a}{b} + x \right )}}{b^{2}} + \frac {c \log {\left (\frac {a}{b} + x \right )}}{b} + \frac {d x}{b} & \text {for}\: m = -1 \\- \frac {a^{2} d \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {a b c m \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {2 a b c \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {a b d m x \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {b^{2} c m x \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {2 b^{2} c x \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {b^{2} d m x^{2} \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} + \frac {b^{2} d x^{2} \left (a + b x\right )^{m}}{b^{2} m^{2} + 3 b^{2} m + 2 b^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c),x)

[Out]

Piecewise((a**m*(c*x + d*x**2/2), Eq(b, 0)), (a*d*log(a/b + x)/(a*b**2 + b**3*x) + a*d/(a*b**2 + b**3*x) - b*c
/(a*b**2 + b**3*x) + b*d*x*log(a/b + x)/(a*b**2 + b**3*x), Eq(m, -2)), (-a*d*log(a/b + x)/b**2 + c*log(a/b + x
)/b + d*x/b, Eq(m, -1)), (-a**2*d*(a + b*x)**m/(b**2*m**2 + 3*b**2*m + 2*b**2) + a*b*c*m*(a + b*x)**m/(b**2*m*
*2 + 3*b**2*m + 2*b**2) + 2*a*b*c*(a + b*x)**m/(b**2*m**2 + 3*b**2*m + 2*b**2) + a*b*d*m*x*(a + b*x)**m/(b**2*
m**2 + 3*b**2*m + 2*b**2) + b**2*c*m*x*(a + b*x)**m/(b**2*m**2 + 3*b**2*m + 2*b**2) + 2*b**2*c*x*(a + b*x)**m/
(b**2*m**2 + 3*b**2*m + 2*b**2) + b**2*d*m*x**2*(a + b*x)**m/(b**2*m**2 + 3*b**2*m + 2*b**2) + b**2*d*x**2*(a
+ b*x)**m/(b**2*m**2 + 3*b**2*m + 2*b**2), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (46) = 92\).
time = 0.00, size = 147, normalized size = 3.20 \begin {gather*} \frac {-a^{2} d \mathrm {e}^{m \ln \left (a+b x\right )}+a b c m \mathrm {e}^{m \ln \left (a+b x\right )}+2 a b c \mathrm {e}^{m \ln \left (a+b x\right )}+a b d m x \mathrm {e}^{m \ln \left (a+b x\right )}+b^{2} c m x \mathrm {e}^{m \ln \left (a+b x\right )}+2 b^{2} c x \mathrm {e}^{m \ln \left (a+b x\right )}+b^{2} d m x^{2} \mathrm {e}^{m \ln \left (a+b x\right )}+b^{2} d x^{2} \mathrm {e}^{m \ln \left (a+b x\right )}}{b^{2} m^{2}+3 b^{2} m+2 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c),x)

[Out]

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

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Mupad [B]
time = 0.48, size = 88, normalized size = 1.91 \begin {gather*} {\left (a+b\,x\right )}^m\,\left (\frac {a\,\left (2\,b\,c-a\,d+b\,c\,m\right )}{b^2\,\left (m^2+3\,m+2\right )}+\frac {x\,\left (2\,b^2\,c+b^2\,c\,m+a\,b\,d\,m\right )}{b^2\,\left (m^2+3\,m+2\right )}+\frac {d\,x^2\,\left (m+1\right )}{m^2+3\,m+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m*(c + d*x),x)

[Out]

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

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