3.118 \(\int (d x)^m (b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=73 \[ \frac {2 b^2 (b+c x) \left (b x+c x^2\right )^{5/2} (d x)^m \left (-\frac {c x}{b}\right )^{-m-\frac {1}{2}} \, _2F_1\left (\frac {7}{2},-m-\frac {5}{2};\frac {9}{2};\frac {c x}{b}+1\right )}{7 c^3 x^2} \]

[Out]

2/7*b^2*(-c*x/b)^(-1/2-m)*(d*x)^m*(c*x+b)*(c*x^2+b*x)^(5/2)*hypergeom([7/2, -5/2-m],[9/2],c*x/b+1)/c^3/x^2

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Rubi [A]  time = 0.03, antiderivative size = 73, 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 = {674, 67, 65} \[ \frac {2 b^2 (b+c x) \left (b x+c x^2\right )^{5/2} (d x)^m \left (-\frac {c x}{b}\right )^{-m-\frac {1}{2}} \, _2F_1\left (\frac {7}{2},-m-\frac {5}{2};\frac {9}{2};\frac {c x}{b}+1\right )}{7 c^3 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(b*x + c*x^2)^(5/2),x]

[Out]

(2*b^2*(-((c*x)/b))^(-1/2 - m)*(d*x)^m*(b + c*x)*(b*x + c*x^2)^(5/2)*Hypergeometric2F1[7/2, -5/2 - m, 9/2, 1 +
(c*x)/b])/(7*c^3*x^2)

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])

Rule 67

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 674

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 )^{5/2} \, dx &=\frac {\left (x^{-\frac {5}{2}-m} (d x)^m \left (b x+c x^2\right )^{5/2}\right ) \int x^{\frac {5}{2}+m} (b+c x)^{5/2} \, dx}{(b+c x)^{5/2}}\\ &=\frac {\left (b^2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m \left (b x+c x^2\right )^{5/2}\right ) \int \left (-\frac {c x}{b}\right )^{\frac {5}{2}+m} (b+c x)^{5/2} \, dx}{c^2 x^2 (b+c x)^{5/2}}\\ &=\frac {2 b^2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m (b+c x) \left (b x+c x^2\right )^{5/2} \, _2F_1\left (\frac {7}{2},-\frac {5}{2}-m;\frac {9}{2};1+\frac {c x}{b}\right )}{7 c^3 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 70, normalized size = 0.96 \[ \frac {2 b^2 (b+c x)^3 \sqrt {x (b+c x)} (d x)^m \left (-\frac {c x}{b}\right )^{-m-\frac {1}{2}} \, _2F_1\left (\frac {7}{2},-m-\frac {5}{2};\frac {9}{2};\frac {c x}{b}+1\right )}{7 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*(b*x + c*x^2)^(5/2),x]

[Out]

(2*b^2*(-((c*x)/b))^(-1/2 - m)*(d*x)^m*(b + c*x)^3*Sqrt[x*(b + c*x)]*Hypergeometric2F1[7/2, -5/2 - m, 9/2, 1 +
(c*x)/b])/(7*c^3)

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}\right )} \sqrt {c x^{2} + b x} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + b^2*x^2)*sqrt(c*x^2 + b*x)*(d*x)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x\right )}^{\frac {5}{2}} \left (d x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(d*x)^m, x)

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+b x \right )^{\frac {5}{2}} \left (d x \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(c*x^2+b*x)^(5/2),x)

[Out]

int((d*x)^m*(c*x^2+b*x)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x\right )}^{\frac {5}{2}} \left (d x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(d*x)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,x^2+b\,x\right )}^{5/2}\,{\left (d\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(5/2)*(d*x)^m,x)

[Out]

int((b*x + c*x^2)^(5/2)*(d*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{m} \left (x \left (b + c x\right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d*x)**m*(x*(b + c*x))**(5/2), x)

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