3.2.19 \(\int (d x)^m (b x+c x^2)^{3/2} \, dx\) [119]

Optimal. Leaf size=71 \[ -\frac {2 b \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m (b+c x) \left (b x+c x^2\right )^{3/2} \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-m;\frac {7}{2};1+\frac {c x}{b}\right )}{5 c^2 x} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 71, 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, 69, 67} \begin {gather*} -\frac {2 b (b+c x) \left (b x+c x^2\right )^{3/2} (d x)^m \left (-\frac {c x}{b}\right )^{-m-\frac {1}{2}} \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {c x}{b}+1\right )}{5 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(b*x + c*x^2)^(3/2),x]

[Out]

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

Rule 67

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

Rule 69

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

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

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Mathematica [A]
time = 0.16, size = 60, normalized size = 0.85 \begin {gather*} -\frac {2 \left (-\frac {c x}{b}\right )^{-\frac {5}{2}-m} (d x)^m (x (b+c x))^{5/2} \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-m;\frac {7}{2};1+\frac {c x}{b}\right )}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (d x \right )^{m} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((c*x^2 + b*x)^(3/2)*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,x^2+b\,x\right )}^{3/2}\,{\left (d\,x\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)*(d*x)^m,x)

[Out]

int((b*x + c*x^2)^(3/2)*(d*x)^m, x)

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