### 3.957 $$\int \frac{(d+e x)^m}{(d^2-e^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=60 $-\frac{(d+e x)^m \, _2F_1\left (1,m-3;m-\frac{1}{2};\frac{d+e x}{2 d}\right )}{d e (3-2 m) \left (d^2-e^2 x^2\right )^{3/2}}$

[Out]

-(((d + e*x)^m*Hypergeometric2F1[1, -3 + m, -1/2 + m, (d + e*x)/(2*d)])/(d*e*(3 - 2*m)*(d^2 - e^2*x^2)^(3/2)))

________________________________________________________________________________________

Rubi [A]  time = 0.052316, antiderivative size = 83, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {680, 678, 69} $\frac{2^{m-\frac{3}{2}} (d+e x)^m \left (\frac{e x}{d}+1\right )^{\frac{3}{2}-m} \, _2F_1\left (-\frac{3}{2},\frac{5}{2}-m;-\frac{1}{2};\frac{d-e x}{2 d}\right )}{3 d e \left (d^2-e^2 x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(2^(-3/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(3/2 - m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (d - e*x)/(2*d)])/(3*
d*e*(d^2 - e^2*x^2)^(3/2))

Rule 680

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPart[m]*(d + e*x)^FracPart[m]
)/(1 + (e*x)/d)^FracPart[m], Int[(1 + (e*x)/d)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && EqQ[c*d
^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] || GtQ[d, 0])

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

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 (d^2-e^2 x^2\right )^{5/2}} \, dx &=\left ((d+e x)^m \left (1+\frac{e x}{d}\right )^{-m}\right ) \int \frac{\left (1+\frac{e x}{d}\right )^m}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac{\left ((d+e x)^m \left (1+\frac{e x}{d}\right )^{\frac{3}{2}-m} \left (d^2-d e x\right )^{3/2}\right ) \int \frac{\left (1+\frac{e x}{d}\right )^{-\frac{5}{2}+m}}{\left (d^2-d e x\right )^{5/2}} \, dx}{\left (d^2-e^2 x^2\right )^{3/2}}\\ &=\frac{2^{-\frac{3}{2}+m} (d+e x)^m \left (1+\frac{e x}{d}\right )^{\frac{3}{2}-m} \, _2F_1\left (-\frac{3}{2},\frac{5}{2}-m;-\frac{1}{2};\frac{d-e x}{2 d}\right )}{3 d e \left (d^2-e^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.111728, size = 92, normalized size = 1.53 $\frac{2^{m-\frac{3}{2}} (d+e x)^m \left (\frac{e x}{d}+1\right )^{\frac{1}{2}-m} \, _2F_1\left (-\frac{3}{2},\frac{5}{2}-m;-\frac{1}{2};\frac{d-e x}{2 d}\right )}{\left (3 d^3 e-3 d^2 e^2 x\right ) \sqrt{d^2-e^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^m/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(2^(-3/2 + m)*(d + e*x)^m*(1 + (e*x)/d)^(1/2 - m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (d - e*x)/(2*d)])/((3
*d^3*e - 3*d^2*e^2*x)*Sqrt[d^2 - e^2*x^2])

________________________________________________________________________________________

Maple [F]  time = 0.484, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m/(-e^2*x^2+d^2)^(5/2),x)

[Out]

int((e*x+d)^m/(-e^2*x^2+d^2)^(5/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/(-e^2*x^2 + d^2)^(5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + d\right )}^{m}}{e^{6} x^{6} - 3 \, d^{2} e^{4} x^{4} + 3 \, d^{4} e^{2} x^{2} - d^{6}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-e^2*x^2 + d^2)*(e*x + d)^m/(e^6*x^6 - 3*d^2*e^4*x^4 + 3*d^4*e^2*x^2 - d^6), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral((d + e*x)**m/(-(-d + e*x)*(d + e*x))**(5/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^m/(-e^2*x^2 + d^2)^(5/2), x)