3.10.44 \(\int \frac {(d+e x)^m (a+b x+c x^2)}{(e+f x)^{3/2}} \, dx\) [944]

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

[Out]

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

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Rubi [A]
time = 0.23, 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 = {963, 81, 72, 71} \begin {gather*} -\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)} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 81

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 963

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]

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*}

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Mathematica [A]
time = 0.52, size = 171, normalized size = 0.72 \begin {gather*} \frac {2 (d+e x)^m \left (\frac {f (d+e x)}{-e^2+d f}\right )^{-m} \left (-3 \left (c e^2+f (-b e+a f)\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}} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )}{\left (f x +e \right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, 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((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)*(x*e + d)^m/(f*x + e)^(3/2), 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((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)*(x*e + d)^m/(f^2*x^2 + 2*f*x*e + e^2), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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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((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)*(x*e + d)^m/(f*x + e)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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