3.8.75 \(\int (d+e x)^m (f+g x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)^{-m} \, dx\) [775]

Optimal. Leaf size=105 \[ \frac {2 \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m (d+e x)^m (f+g x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, _2F_1\left (\frac {5}{2},m;\frac {7}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{5 g} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 105, normalized size of antiderivative = 1.00, 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)^{5/2} (d+e x)^m \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m \, _2F_1\left (\frac {5}{2},m;\frac {7}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{5 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-((g*(a*e + c*d*x))/(c*d*f - a*e*g)))^m*(d + e*x)^m*(f + g*x)^(5/2)*Hypergeometric2F1[5/2, m, 7/2, (c*d*(f
+ g*x))/(c*d*f - a*e*g)])/(5*g*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^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 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 (d+e x)^m (f+g x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx &=\left ((a e+c d x)^m (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}\right ) \int (a e+c d x)^{-m} (f+g x)^{3/2} \, dx\\ &=\left (\left (\frac {g (a e+c d x)}{-c d f+a e g}\right )^m (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}\right ) \int (f+g x)^{3/2} \left (-\frac {a e g}{c d f-a e g}-\frac {c d g x}{c d f-a e g}\right )^{-m} \, dx\\ &=\frac {2 \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m (d+e x)^m (f+g x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, _2F_1\left (\frac {5}{2},m;\frac {7}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{5 g}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 93, normalized size = 0.89 \begin {gather*} \frac {2 \left (\frac {g (a e+c d x)}{-c d f+a e g}\right )^m (d+e x)^m ((a e+c d x) (d+e x))^{-m} (f+g x)^{5/2} \, _2F_1\left (\frac {5}{2},m;\frac {7}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{5 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4371 deep

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

[Out]

integrate((g*x + f)^(3/2)*(x*e + d)^m/(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^m, 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 )}^{3/2}\,{\left (d+e\,x\right )}^m}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^m} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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