3.32.86 \(\int \frac {(A+B x) (d+e x)^m}{(a+b x)^2} \, dx\) [3186]

Optimal. Leaf size=112 \[ -\frac {(A b-a B) (d+e x)^{1+m}}{b (b d-a e) (a+b x)}+\frac {(a B e (1+m)-b (B d+A e m)) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{b (b d-a e)^2 (1+m)} \]

[Out]

-(A*b-B*a)*(e*x+d)^(1+m)/b/(-a*e+b*d)/(b*x+a)+(a*B*e*(1+m)-b*(A*e*m+B*d))*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+
m],b*(e*x+d)/(-a*e+b*d))/b/(-a*e+b*d)^2/(1+m)

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Rubi [A]
time = 0.04, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {79, 70} \begin {gather*} \frac {(d+e x)^{m+1} (a B e (m+1)-b (A e m+B d)) \, _2F_1\left (1,m+1;m+2;\frac {b (d+e x)}{b d-a e}\right )}{b (m+1) (b d-a e)^2}-\frac {(A b-a B) (d+e x)^{m+1}}{b (a+b x) (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^m)/(a + b*x)^2,x]

[Out]

-(((A*b - a*B)*(d + e*x)^(1 + m))/(b*(b*d - a*e)*(a + b*x))) + ((a*B*e*(1 + m) - b*(B*d + A*e*m))*(d + e*x)^(1
+ m)*Hypergeometric2F1[1, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(b*(b*d - a*e)^2*(1 + m))

Rule 70

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

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^m}{(a+b x)^2} \, dx &=-\frac {(A b-a B) (d+e x)^{1+m}}{b (b d-a e) (a+b x)}-\frac {(a B e (1+m)-b (B d+A e m)) \int \frac {(d+e x)^m}{a+b x} \, dx}{b (b d-a e)}\\ &=-\frac {(A b-a B) (d+e x)^{1+m}}{b (b d-a e) (a+b x)}+\frac {(a B e (1+m)-b (B d+A e m)) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{b (b d-a e)^2 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 99, normalized size = 0.88 \begin {gather*} \frac {(d+e x)^{1+m} \left (-\frac {(A b-a B) (b d-a e)}{a+b x}+\frac {(a B e (1+m)-b (B d+A e m)) \, _2F_1\left (1,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{1+m}\right )}{b (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(-(((A*b - a*B)*(b*d - a*e))/(a + b*x)) + ((a*B*e*(1 + m) - b*(B*d + A*e*m))*Hypergeometric
2F1[1, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(1 + m)))/(b*(b*d - a*e)^2)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (b x +a \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^m/(b*x+a)^2,x)

[Out]

int((B*x+A)*(e*x+d)^m/(b*x+a)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(x*e + d)^m/(b*x + a)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x + A)*(x*e + d)^m/(b^2*x^2 + 2*a*b*x + a^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )^{m}}{\left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**m/(b*x+a)**2,x)

[Out]

Integral((A + B*x)*(d + e*x)**m/(a + b*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(x*e + d)^m/(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^m)/(a + b*x)^2,x)

[Out]

int(((A + B*x)*(d + e*x)^m)/(a + b*x)^2, x)

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