3.2008 $$\int \frac{1}{(d+e x)^{7/2} (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx$$

Optimal. Leaf size=186 $\frac{2 c^3 d^3}{\sqrt{d+e x} \left (c d^2-a e^2\right )^4}+\frac{2 c^2 d^2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}-\frac{2 c^{7/2} d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}}+\frac{2 c d}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2}+\frac{2}{7 (d+e x)^{7/2} \left (c d^2-a e^2\right )}$

[Out]

2/(7*(c*d^2 - a*e^2)*(d + e*x)^(7/2)) + (2*c*d)/(5*(c*d^2 - a*e^2)^2*(d + e*x)^(5/2)) + (2*c^2*d^2)/(3*(c*d^2
- a*e^2)^3*(d + e*x)^(3/2)) + (2*c^3*d^3)/((c*d^2 - a*e^2)^4*Sqrt[d + e*x]) - (2*c^(7/2)*d^(7/2)*ArcTanh[(Sqrt
[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(9/2)

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Rubi [A]  time = 0.163654, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {626, 51, 63, 208} $\frac{2 c^3 d^3}{\sqrt{d+e x} \left (c d^2-a e^2\right )^4}+\frac{2 c^2 d^2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}-\frac{2 c^{7/2} d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}}+\frac{2 c d}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2}+\frac{2}{7 (d+e x)^{7/2} \left (c d^2-a e^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(7/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

2/(7*(c*d^2 - a*e^2)*(d + e*x)^(7/2)) + (2*c*d)/(5*(c*d^2 - a*e^2)^2*(d + e*x)^(5/2)) + (2*c^2*d^2)/(3*(c*d^2
- a*e^2)^3*(d + e*x)^(3/2)) + (2*c^3*d^3)/((c*d^2 - a*e^2)^4*Sqrt[d + e*x]) - (2*c^(7/2)*d^(7/2)*ArcTanh[(Sqrt
[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(9/2)

Rule 626

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

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^{7/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx &=\int \frac{1}{(a e+c d x) (d+e x)^{9/2}} \, dx\\ &=\frac{2}{7 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}+\frac{(c d) \int \frac{1}{(a e+c d x) (d+e x)^{7/2}} \, dx}{c d^2-a e^2}\\ &=\frac{2}{7 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}+\frac{2 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}+\frac{\left (c^2 d^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{\left (c d^2-a e^2\right )^2}\\ &=\frac{2}{7 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}+\frac{2 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}+\frac{2 c^2 d^2}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}+\frac{\left (c^3 d^3\right ) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{\left (c d^2-a e^2\right )^3}\\ &=\frac{2}{7 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}+\frac{2 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}+\frac{2 c^2 d^2}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}+\frac{2 c^3 d^3}{\left (c d^2-a e^2\right )^4 \sqrt{d+e x}}+\frac{\left (c^4 d^4\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{\left (c d^2-a e^2\right )^4}\\ &=\frac{2}{7 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}+\frac{2 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}+\frac{2 c^2 d^2}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}+\frac{2 c^3 d^3}{\left (c d^2-a e^2\right )^4 \sqrt{d+e x}}+\frac{\left (2 c^4 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c d^2}{e}+a e+\frac{c d x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{e \left (c d^2-a e^2\right )^4}\\ &=\frac{2}{7 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}+\frac{2 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}+\frac{2 c^2 d^2}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}+\frac{2 c^3 d^3}{\left (c d^2-a e^2\right )^4 \sqrt{d+e x}}-\frac{2 c^{7/2} d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0138336, size = 57, normalized size = 0.31 $\frac{2 \, _2F_1\left (-\frac{7}{2},1;-\frac{5}{2};\frac{c d (d+e x)}{c d^2-a e^2}\right )}{7 (d+e x)^{7/2} \left (c d^2-a e^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(7/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

(2*Hypergeometric2F1[-7/2, 1, -5/2, (c*d*(d + e*x))/(c*d^2 - a*e^2)])/(7*(c*d^2 - a*e^2)*(d + e*x)^(7/2))

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Maple [A]  time = 0.199, size = 175, normalized size = 0.9 \begin{align*} -{\frac{2}{7\,a{e}^{2}-7\,c{d}^{2}} \left ( ex+d \right ) ^{-{\frac{7}{2}}}}-{\frac{2\,{c}^{2}{d}^{2}}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,cd}{5\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}+2\,{\frac{{c}^{3}{d}^{3}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}\sqrt{ex+d}}}+2\,{\frac{{c}^{4}{d}^{4}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}\arctan \left ({\frac{\sqrt{ex+d}cd}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x)

[Out]

-2/7/(a*e^2-c*d^2)/(e*x+d)^(7/2)-2/3*c^2*d^2/(a*e^2-c*d^2)^3/(e*x+d)^(3/2)+2/5*c*d/(a*e^2-c*d^2)^2/(e*x+d)^(5/
2)+2*c^3*d^3/(a*e^2-c*d^2)^4/(e*x+d)^(1/2)+2*c^4*d^4/(a*e^2-c*d^2)^4/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^
(1/2)*c*d/((a*e^2-c*d^2)*c*d)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.03236, size = 2345, normalized size = 12.61 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

[1/105*(105*(c^3*d^3*e^4*x^4 + 4*c^3*d^4*e^3*x^3 + 6*c^3*d^5*e^2*x^2 + 4*c^3*d^6*e*x + c^3*d^7)*sqrt(c*d/(c*d^
2 - a*e^2))*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(c*d/(c*d^2 - a*e^2)))/(c*d*x
+ a*e)) + 2*(105*c^3*d^3*e^3*x^3 + 176*c^3*d^6 - 122*a*c^2*d^4*e^2 + 66*a^2*c*d^2*e^4 - 15*a^3*e^6 + 35*(10*c
^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 + 7*(58*c^3*d^5*e - 16*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^4*d
^12 - 4*a*c^3*d^10*e^2 + 6*a^2*c^2*d^8*e^4 - 4*a^3*c*d^6*e^6 + a^4*d^4*e^8 + (c^4*d^8*e^4 - 4*a*c^3*d^6*e^6 +
6*a^2*c^2*d^4*e^8 - 4*a^3*c*d^2*e^10 + a^4*e^12)*x^4 + 4*(c^4*d^9*e^3 - 4*a*c^3*d^7*e^5 + 6*a^2*c^2*d^5*e^7 -
4*a^3*c*d^3*e^9 + a^4*d*e^11)*x^3 + 6*(c^4*d^10*e^2 - 4*a*c^3*d^8*e^4 + 6*a^2*c^2*d^6*e^6 - 4*a^3*c*d^4*e^8 +
a^4*d^2*e^10)*x^2 + 4*(c^4*d^11*e - 4*a*c^3*d^9*e^3 + 6*a^2*c^2*d^7*e^5 - 4*a^3*c*d^5*e^7 + a^4*d^3*e^9)*x), -
2/105*(105*(c^3*d^3*e^4*x^4 + 4*c^3*d^4*e^3*x^3 + 6*c^3*d^5*e^2*x^2 + 4*c^3*d^6*e*x + c^3*d^7)*sqrt(-c*d/(c*d^
2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2)) - (105*c^3*d^3
*e^3*x^3 + 176*c^3*d^6 - 122*a*c^2*d^4*e^2 + 66*a^2*c*d^2*e^4 - 15*a^3*e^6 + 35*(10*c^3*d^4*e^2 - a*c^2*d^2*e^
4)*x^2 + 7*(58*c^3*d^5*e - 16*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^4*d^12 - 4*a*c^3*d^10*e^2 +
6*a^2*c^2*d^8*e^4 - 4*a^3*c*d^6*e^6 + a^4*d^4*e^8 + (c^4*d^8*e^4 - 4*a*c^3*d^6*e^6 + 6*a^2*c^2*d^4*e^8 - 4*a^3
*c*d^2*e^10 + a^4*e^12)*x^4 + 4*(c^4*d^9*e^3 - 4*a*c^3*d^7*e^5 + 6*a^2*c^2*d^5*e^7 - 4*a^3*c*d^3*e^9 + a^4*d*e
^11)*x^3 + 6*(c^4*d^10*e^2 - 4*a*c^3*d^8*e^4 + 6*a^2*c^2*d^6*e^6 - 4*a^3*c*d^4*e^8 + a^4*d^2*e^10)*x^2 + 4*(c^
4*d^11*e - 4*a*c^3*d^9*e^3 + 6*a^2*c^2*d^7*e^5 - 4*a^3*c*d^5*e^7 + a^4*d^3*e^9)*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(1/(e*x+d)**(7/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)

[Out]

Timed out

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Giac [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(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")

[Out]

Timed out