3.74 \(\int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx\)

Optimal. Leaf size=107 \[ \frac {8 a^2 c^3 (e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{e (m+1)}-\frac {7 a^2 c^3 (e x)^{m+1}}{e (m+1)}+\frac {4 a b c^3 (e x)^{m+2}}{e^2 (m+2)}-\frac {b^2 c^3 (e x)^{m+3}}{e^3 (m+3)} \]

[Out]

-7*a^2*c^3*(e*x)^(1+m)/e/(1+m)+4*a*b*c^3*(e*x)^(2+m)/e^2/(2+m)-b^2*c^3*(e*x)^(3+m)/e^3/(3+m)+8*a^2*c^3*(e*x)^(
1+m)*hypergeom([1, 1+m],[2+m],-b*x/a)/e/(1+m)

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Rubi [A]  time = 0.12, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {88, 64, 43} \[ \frac {8 a^2 c^3 (e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{e (m+1)}-\frac {7 a^2 c^3 (e x)^{m+1}}{e (m+1)}+\frac {4 a b c^3 (e x)^{m+2}}{e^2 (m+2)}-\frac {b^2 c^3 (e x)^{m+3}}{e^3 (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(a*c - b*c*x)^3)/(a + b*x),x]

[Out]

(-7*a^2*c^3*(e*x)^(1 + m))/(e*(1 + m)) + (4*a*b*c^3*(e*x)^(2 + m))/(e^2*(2 + m)) - (b^2*c^3*(e*x)^(3 + m))/(e^
3*(3 + m)) + (8*a^2*c^3*(e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(e*(1 + m))

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])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {(e x)^m (a c-b c x)^3}{a+b x} \, dx &=\int \left (-4 a^2 c^3 (e x)^m+\frac {8 a^3 c^3 (e x)^m}{a+b x}-2 a c^2 (e x)^m (a c-b c x)-c (e x)^m (a c-b c x)^2\right ) \, dx\\ &=-\frac {4 a^2 c^3 (e x)^{1+m}}{e (1+m)}-c \int (e x)^m (a c-b c x)^2 \, dx-\left (2 a c^2\right ) \int (e x)^m (a c-b c x) \, dx+\left (8 a^3 c^3\right ) \int \frac {(e x)^m}{a+b x} \, dx\\ &=-\frac {4 a^2 c^3 (e x)^{1+m}}{e (1+m)}+\frac {8 a^2 c^3 (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{e (1+m)}-c \int \left (a^2 c^2 (e x)^m-\frac {2 a b c^2 (e x)^{1+m}}{e}+\frac {b^2 c^2 (e x)^{2+m}}{e^2}\right ) \, dx-\left (2 a c^2\right ) \int \left (a c (e x)^m-\frac {b c (e x)^{1+m}}{e}\right ) \, dx\\ &=-\frac {7 a^2 c^3 (e x)^{1+m}}{e (1+m)}+\frac {4 a b c^3 (e x)^{2+m}}{e^2 (2+m)}-\frac {b^2 c^3 (e x)^{3+m}}{e^3 (3+m)}+\frac {8 a^2 c^3 (e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{e (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 69, normalized size = 0.64 \[ c^3 x (e x)^m \left (\frac {8 a^2 \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{m+1}-\frac {7 a^2}{m+1}+\frac {4 a b x}{m+2}-\frac {b^2 x^2}{m+3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(a*c - b*c*x)^3)/(a + b*x),x]

[Out]

c^3*x*(e*x)^m*((-7*a^2)/(1 + m) + (4*a*b*x)/(2 + m) - (b^2*x^2)/(3 + m) + (8*a^2*Hypergeometric2F1[1, 1 + m, 2
+ m, -((b*x)/a)])/(1 + m))

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 3 \, a^{2} b c^{3} x - a^{3} c^{3}\right )} \left (e x\right )^{m}}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(-b*c*x+a*c)^3/(b*x+a),x, algorithm="fricas")

[Out]

integral(-(b^3*c^3*x^3 - 3*a*b^2*c^3*x^2 + 3*a^2*b*c^3*x - a^3*c^3)*(e*x)^m/(b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b c x - a c\right )}^{3} \left (e x\right )^{m}}{b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(-b*c*x+a*c)^3/(b*x+a),x, algorithm="giac")

[Out]

integrate(-(b*c*x - a*c)^3*(e*x)^m/(b*x + a), x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b c x +a c \right )^{3} \left (e x \right )^{m}}{b x +a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(-b*c*x+a*c)^3/(b*x+a),x)

[Out]

int((e*x)^m*(-b*c*x+a*c)^3/(b*x+a),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (b c x - a c\right )}^{3} \left (e x\right )^{m}}{b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(-b*c*x+a*c)^3/(b*x+a),x, algorithm="maxima")

[Out]

-integrate((b*c*x - a*c)^3*(e*x)^m/(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a\,c-b\,c\,x\right )}^3\,{\left (e\,x\right )}^m}{a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*c - b*c*x)^3*(e*x)^m)/(a + b*x),x)

[Out]

int(((a*c - b*c*x)^3*(e*x)^m)/(a + b*x), x)

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sympy [C]  time = 6.52, size = 340, normalized size = 3.18 \[ \frac {a^{2} c^{3} e^{m} m x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac {a^{2} c^{3} e^{m} x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} - \frac {3 a b c^{3} e^{m} m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} - \frac {6 a b c^{3} e^{m} x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac {3 b^{2} c^{3} e^{m} m x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{\Gamma \left (m + 4\right )} + \frac {9 b^{2} c^{3} e^{m} x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{\Gamma \left (m + 4\right )} - \frac {b^{3} c^{3} e^{m} m x^{4} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} - \frac {4 b^{3} c^{3} e^{m} x^{4} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(-b*c*x+a*c)**3/(b*x+a),x)

[Out]

a**2*c**3*e**m*m*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) + a**2*c**3*e**m*x
*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/gamma(m + 2) - 3*a*b*c**3*e**m*m*x**2*x**m*lerchp
hi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/gamma(m + 3) - 6*a*b*c**3*e**m*x**2*x**m*lerchphi(b*x*exp_pol
ar(I*pi)/a, 1, m + 2)*gamma(m + 2)/gamma(m + 3) + 3*b**2*c**3*e**m*m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a,
1, m + 3)*gamma(m + 3)/gamma(m + 4) + 9*b**2*c**3*e**m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*ga
mma(m + 3)/gamma(m + 4) - b**3*c**3*e**m*m*x**4*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a
*gamma(m + 5)) - 4*b**3*c**3*e**m*x**4*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a*gamma(m
+ 5))

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