3.944 \(\int \frac{(d+e x)^m (a+b x+c x^2)}{(e+f x)^{3/2}} \, dx\)

Optimal. Leaf size=237 \[ \frac{2 \sqrt{e+f x} (d+e x)^m \left (-\frac{f (d+e x)}{e^2-d f}\right )^{-m} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{e (e+f x)}{e^2-d f}\right ) \left (c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )-e f (2 m+3) \left (a e f (2 m+1)+b \left (d f-2 e^2 (m+1)\right )\right )\right )}{e f^3 (2 m+3) \left (e^2-d f\right )}+\frac{2 (d+e x)^{m+1} \left (a+\frac{e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 c \sqrt{e+f x} (d+e x)^{m+1}}{e f^2 (2 m+3)} \]

[Out]

(2*(a + (e*(c*e - b*f))/f^2)*(d + e*x)^(1 + m))/((e^2 - d*f)*Sqrt[e + f*x]) + (2*c*(d + e*x)^(1 + m)*Sqrt[e +
f*x])/(e*f^2*(3 + 2*m)) + (2*(c*(d^2*f^2 + 4*d*e^2*f*(1 + m) - 4*e^4*(2 + 3*m + m^2)) - e*f*(3 + 2*m)*(a*e*f*(
1 + 2*m) + b*(d*f - 2*e^2*(1 + m))))*(d + e*x)^m*Sqrt[e + f*x]*Hypergeometric2F1[1/2, -m, 3/2, (e*(e + f*x))/(
e^2 - d*f)])/(e*f^3*(e^2 - d*f)*(3 + 2*m)*(-((f*(d + e*x))/(e^2 - d*f)))^m)

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Rubi [A]  time = 0.348564, antiderivative size = 230, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {949, 80, 70, 69} \[ -\frac{2 \sqrt{e+f x} (d+e x)^m \left (-\frac{f (d+e x)}{e^2-d f}\right )^{-m} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{e (e+f x)}{e^2-d f}\right ) \left (f \left (a e f (2 m+1)+b d f-2 b e^2 (m+1)\right )-\frac{c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )}{e (2 m+3)}\right )}{f^3 \left (e^2-d f\right )}+\frac{2 (d+e x)^{m+1} \left (a+\frac{e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 c \sqrt{e+f x} (d+e x)^{m+1}}{e f^2 (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + (e*(c*e - b*f))/f^2)*(d + e*x)^(1 + m))/((e^2 - d*f)*Sqrt[e + f*x]) + (2*c*(d + e*x)^(1 + m)*Sqrt[e +
f*x])/(e*f^2*(3 + 2*m)) - (2*(f*(b*d*f - 2*b*e^2*(1 + m) + a*e*f*(1 + 2*m)) - (c*(d^2*f^2 + 4*d*e^2*f*(1 + m)
- 4*e^4*(2 + 3*m + m^2)))/(e*(3 + 2*m)))*(d + e*x)^m*Sqrt[e + f*x]*Hypergeometric2F1[1/2, -m, 3/2, (e*(e + f*x
))/(e^2 - d*f)])/(f^3*(e^2 - d*f)*(-((f*(d + e*x))/(e^2 - d*f)))^m)

Rule 949

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[(R*(d + e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 70

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

Rule 69

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 \int \frac{(d+e x)^m \left (\frac{c \left (d e f-2 e^3 (1+m)\right )-f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )}{2 f^2}-\frac{1}{2} c \left (d-\frac{e^2}{f}\right ) x\right )}{\sqrt{e+f x}} \, dx}{e^2-d f}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 c (d+e x)^{1+m} \sqrt{e+f x}}{e f^2 (3+2 m)}-\frac{\left (f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )-\frac{c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )}{e (3+2 m)}\right ) \int \frac{(d+e x)^m}{\sqrt{e+f x}} \, dx}{f^2 \left (e^2-d f\right )}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 c (d+e x)^{1+m} \sqrt{e+f x}}{e f^2 (3+2 m)}-\frac{\left (\left (f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )-\frac{c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )}{e (3+2 m)}\right ) (d+e x)^m \left (\frac{f (d+e x)}{-e^2+d f}\right )^{-m}\right ) \int \frac{\left (-\frac{d f}{e^2-d f}-\frac{e f x}{e^2-d f}\right )^m}{\sqrt{e+f x}} \, dx}{f^2 \left (e^2-d f\right )}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 c (d+e x)^{1+m} \sqrt{e+f x}}{e f^2 (3+2 m)}-\frac{2 \left (f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )-\frac{c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )}{e (3+2 m)}\right ) (d+e x)^m \left (-\frac{f (d+e x)}{e^2-d f}\right )^{-m} \sqrt{e+f x} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{e (e+f x)}{e^2-d f}\right )}{f^3 \left (e^2-d f\right )}\\ \end{align*}

Mathematica [A]  time = 0.291858, size = 171, normalized size = 0.72 \[ \frac{2 (d+e x)^m \left (\frac{f (d+e x)}{d f-e^2}\right )^{-m} \left (-3 \left (f (a f-b e)+c e^2\right ) \, _2F_1\left (-\frac{1}{2},-m;\frac{1}{2};\frac{e (e+f x)}{e^2-d f}\right )-(e+f x) \left ((6 c e-3 b f) \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{e (e+f x)}{e^2-d f}\right )-c (e+f x) \, _2F_1\left (\frac{3}{2},-m;\frac{5}{2};\frac{e (e+f x)}{e^2-d f}\right )\right )\right )}{3 f^3 \sqrt{e+f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^m*(-3*(c*e^2 + f*(-(b*e) + a*f))*Hypergeometric2F1[-1/2, -m, 1/2, (e*(e + f*x))/(e^2 - d*f)] - (e
+ f*x)*((6*c*e - 3*b*f)*Hypergeometric2F1[1/2, -m, 3/2, (e*(e + f*x))/(e^2 - d*f)] - c*(e + f*x)*Hypergeometr
ic2F1[3/2, -m, 5/2, (e*(e + f*x))/(e^2 - d*f)])))/(3*f^3*((f*(d + e*x))/(-e^2 + d*f))^m*Sqrt[e + f*x])

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Maple [F]  time = 0.628, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) \left ( fx+e \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}{\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}{\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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