3.8.65 \(\int \frac {(f+g x)^n (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{3/2}} \, dx\) [765]

Optimal. Leaf size=104 \[ -\frac {(a e+c d x) (f+g x)^{1+n} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2} \, _2F_1\left (1,\frac {7}{2}+n;2+n;\frac {c d (f+g x)}{c d f-a e g}\right )}{(c d f-a e g) (1+n) (d+e x)^{3/2}} \]

[Out]

-(c*d*x+a*e)*(g*x+f)^(1+n)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*hypergeom([1, 7/2+n],[2+n],c*d*(g*x+f)/(-a*
e*g+c*d*f))/(-a*e*g+c*d*f)/(1+n)/(e*x+d)^(3/2)

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Rubi [A]
time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {905, 72, 71} \begin {gather*} \frac {2 (f+g x)^n (a e+c d x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {5}{2},-n;\frac {7}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{5 c d \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*e + c*d*x)^2*(f + g*x)^n*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]*Hypergeometric2F1[5/2, -n, 7/2, -((
g*(a*e + c*d*x))/(c*d*f - a*e*g))])/(5*c*d*Sqrt[d + e*x]*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

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 905

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(f+g x)^n \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \int (a e+c d x)^{3/2} (f+g x)^n \, dx}{\sqrt {a e+c d x} \sqrt {d+e x}}\\ &=\frac {\left ((f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}\right ) \int (a e+c d x)^{3/2} \left (\frac {c d f}{c d f-a e g}+\frac {c d g x}{c d f-a e g}\right )^n \, dx}{\sqrt {a e+c d x} \sqrt {d+e x}}\\ &=\frac {2 (a e+c d x)^2 (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, _2F_1\left (\frac {5}{2},-n;\frac {7}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{5 c d \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 100, normalized size = 0.96 \begin {gather*} \frac {2 ((a e+c d x) (d+e x))^{5/2} (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {5}{2},-n;\frac {7}{2};\frac {g (a e+c d x)}{-c d f+a e g}\right )}{5 c d (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(f + g*x)^n*Hypergeometric2F1[5/2, -n, 7/2, (g*(a*e + c*d*x))/(-(c*d*f) + a
*e*g)])/(5*c*d*(d + e*x)^(5/2)*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(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((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**n*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/(e*x+d)**(3/2),x)

[Out]

Timed out

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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((g*x+f)^n*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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