3.2.32 \(\int (d x)^{3/2} (b x+c x^2)^p \, dx\) [132]

Optimal. Leaf size=61 \[ \frac {2 x (d x)^{3/2} \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,\frac {5}{2}+p;\frac {7}{2}+p;-\frac {c x}{b}\right )}{5+2 p} \]

[Out]

2*x*(d*x)^(3/2)*(c*x^2+b*x)^p*hypergeom([-p, 5/2+p],[7/2+p],-c*x/b)/(5+2*p)/((c*x/b+1)^p)

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Rubi [A]
time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {688, 68, 66} \begin {gather*} \frac {2 x (d x)^{3/2} \left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,p+\frac {5}{2};p+\frac {7}{2};-\frac {c x}{b}\right )}{2 p+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*(b*x + c*x^2)^p,x]

[Out]

(2*x*(d*x)^(3/2)*(b*x + c*x^2)^p*Hypergeometric2F1[-p, 5/2 + p, 7/2 + p, -((c*x)/b)])/((5 + 2*p)*(1 + (c*x)/b)
^p)

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 68

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

Rule 688

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*
(b + c*x)^p)), Int[x^(m + p)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, m}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int (d x)^{3/2} \left (b x+c x^2\right )^p \, dx &=\left (x^{-\frac {3}{2}-p} (d x)^{3/2} (b+c x)^{-p} \left (b x+c x^2\right )^p\right ) \int x^{\frac {3}{2}+p} (b+c x)^p \, dx\\ &=\left (x^{-\frac {3}{2}-p} (d x)^{3/2} \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p\right ) \int x^{\frac {3}{2}+p} \left (1+\frac {c x}{b}\right )^p \, dx\\ &=\frac {2 x (d x)^{3/2} \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,\frac {5}{2}+p;\frac {7}{2}+p;-\frac {c x}{b}\right )}{5+2 p}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 58, normalized size = 0.95 \begin {gather*} \frac {x (d x)^{3/2} (x (b+c x))^p \left (1+\frac {c x}{b}\right )^{-p} \, _2F_1\left (-p,\frac {5}{2}+p;\frac {7}{2}+p;-\frac {c x}{b}\right )}{\frac {5}{2}+p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^(3/2)*(x*(b + c*x))^p*Hypergeometric2F1[-p, 5/2 + p, 7/2 + p, -((c*x)/b)])/((5/2 + p)*(1 + (c*x)/b)^p
)

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \left (d x \right )^{\frac {3}{2}} \left (c \,x^{2}+b x \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{\frac {3}{2}} \left (x \left (b + c x\right )\right )^{p}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (c\,x^2+b\,x\right )}^p\,{\left (d\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^p*(d*x)^(3/2),x)

[Out]

int((b*x + c*x^2)^p*(d*x)^(3/2), x)

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