3.2852 \(\int \frac{(a+b (c+d x)^2)^p}{c+d x} \, dx\)

Optimal. Leaf size=52 \[ -\frac{\left (a+b (c+d x)^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b (c+d x)^2}{a}+1\right )}{2 a d (p+1)} \]

[Out]

-((a + b*(c + d*x)^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c + d*x)^2)/a])/(2*a*d*(1 + p))

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Rubi [A]  time = 0.0532383, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {372, 266, 65} \[ -\frac{\left (a+b (c+d x)^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b (c+d x)^2}{a}+1\right )}{2 a d (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c + d*x)^2)^p/(c + d*x),x]

[Out]

-((a + b*(c + d*x)^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c + d*x)^2)/a])/(2*a*d*(1 + p))

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

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

Rubi steps

\begin{align*} \int \frac{\left (a+b (c+d x)^2\right )^p}{c+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,(c+d x)^2\right )}{2 d}\\ &=-\frac{\left (a+b (c+d x)^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b (c+d x)^2}{a}\right )}{2 a d (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0244537, size = 52, normalized size = 1. \[ -\frac{\left (a+b (c+d x)^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b (c+d x)^2}{a}+1\right )}{2 a d (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*(c + d*x)^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c + d*x)^2)/a])/(2*a*d*(1 + p))

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Maple [F]  time = 0.12, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b \left ( dx+c \right ) ^{2} \right ) ^{p}}{dx+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left ({\left (d x + c\right )}^{2} b + a\right )}^{p}}{d x + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{d x + c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p/(d*x + c), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left ({\left (d x + c\right )}^{2} b + a\right )}^{p}}{d x + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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