3.446 \(\int (d+e x)^m (b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=105 \[ \frac {\left (b x+c x^2\right )^{3/2} (d+e x)^{m+1} F_1\left (m+1;-\frac {3}{2},-\frac {3}{2};m+2;\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (m+1) \left (-\frac {e x}{d}\right )^{3/2} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{3/2}} \]

[Out]

(e*x+d)^(1+m)*(c*x^2+b*x)^(3/2)*AppellF1(1+m,-3/2,-3/2,2+m,(e*x+d)/d,c*(e*x+d)/(-b*e+c*d))/e/(1+m)/(-e*x/d)^(3
/2)/(1-c*(e*x+d)/(-b*e+c*d))^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {759, 133} \[ \frac {\left (b x+c x^2\right )^{3/2} (d+e x)^{m+1} F_1\left (m+1;-\frac {3}{2},-\frac {3}{2};m+2;\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (m+1) \left (-\frac {e x}{d}\right )^{3/2} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(b*x + c*x^2)^(3/2)*AppellF1[1 + m, -3/2, -3/2, 2 + m, (d + e*x)/d, (c*(d + e*x))/(c*d - b*
e)])/(e*(1 + m)*(-((e*x)/d))^(3/2)*(1 - (c*(d + e*x))/(c*d - b*e))^(3/2))

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 759

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(a + b*x + c*x^2)^p/(e*(1 - (d + e*x)/(d - (e*(b - q))/(2*c)))^p*(1 - (d + e*x)/(d - (e*(b + q))/(2
*c)))^p), Subst[Int[x^m*Simp[1 - x/(d - (e*(b - q))/(2*c)), x]^p*Simp[1 - x/(d - (e*(b + q))/(2*c)), x]^p, x],
x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int (d+e x)^m \left (b x+c x^2\right )^{3/2} \, dx &=\frac {\left (b x+c x^2\right )^{3/2} \operatorname {Subst}\left (\int x^m \left (1-\frac {x}{d}\right )^{3/2} \left (1-\frac {c x}{c d-b e}\right )^{3/2} \, dx,x,d+e x\right )}{e \left (1-\frac {d+e x}{d}\right )^{3/2} \left (1-\frac {d+e x}{d-\frac {b e}{c}}\right )^{3/2}}\\ &=\frac {(d+e x)^{1+m} \left (b x+c x^2\right )^{3/2} F_1\left (1+m;-\frac {3}{2},-\frac {3}{2};2+m;\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (1+m) \left (-\frac {e x}{d}\right )^{3/2} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 111, normalized size = 1.06 \[ \frac {2 x^2 \sqrt {x (b+c x)} (d+e x)^m \left (\frac {e x}{d}+1\right )^{-m} \left (7 b F_1\left (\frac {5}{2};-\frac {1}{2},-m;\frac {7}{2};-\frac {c x}{b},-\frac {e x}{d}\right )+5 c x F_1\left (\frac {7}{2};-\frac {1}{2},-m;\frac {9}{2};-\frac {c x}{b},-\frac {e x}{d}\right )\right )}{35 \sqrt {\frac {c x}{b}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x^2*Sqrt[x*(b + c*x)]*(d + e*x)^m*(7*b*AppellF1[5/2, -1/2, -m, 7/2, -((c*x)/b), -((e*x)/d)] + 5*c*x*AppellF
1[7/2, -1/2, -m, 9/2, -((c*x)/b), -((e*x)/d)]))/(35*Sqrt[1 + (c*x)/b]*(1 + (e*x)/d)^m)

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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + b x\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.78, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+b x \right )^{\frac {3}{2}} \left (e x +d \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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