3.2066 \(\int \frac{(a+b x) (d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=138 \[ \frac{2 \sqrt{d+e x} (b d-a e)^3}{b^4}+\frac{2 (d+e x)^{3/2} (b d-a e)^2}{3 b^3}+\frac{2 (d+e x)^{5/2} (b d-a e)}{5 b^2}-\frac{2 (b d-a e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2}}+\frac{2 (d+e x)^{7/2}}{7 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.124759, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 50, 63, 208} \[ \frac{2 \sqrt{d+e x} (b d-a e)^3}{b^4}+\frac{2 (d+e x)^{3/2} (b d-a e)^2}{3 b^3}+\frac{2 (d+e x)^{5/2} (b d-a e)}{5 b^2}-\frac{2 (b d-a e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2}}+\frac{2 (d+e x)^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{(a+b x) (d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{(d+e x)^{7/2}}{a+b x} \, dx\\ &=\frac{2 (d+e x)^{7/2}}{7 b}+\frac{(b d-a e) \int \frac{(d+e x)^{5/2}}{a+b x} \, dx}{b}\\ &=\frac{2 (b d-a e) (d+e x)^{5/2}}{5 b^2}+\frac{2 (d+e x)^{7/2}}{7 b}+\frac{(b d-a e)^2 \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{b^2}\\ &=\frac{2 (b d-a e)^2 (d+e x)^{3/2}}{3 b^3}+\frac{2 (b d-a e) (d+e x)^{5/2}}{5 b^2}+\frac{2 (d+e x)^{7/2}}{7 b}+\frac{(b d-a e)^3 \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{b^3}\\ &=\frac{2 (b d-a e)^3 \sqrt{d+e x}}{b^4}+\frac{2 (b d-a e)^2 (d+e x)^{3/2}}{3 b^3}+\frac{2 (b d-a e) (d+e x)^{5/2}}{5 b^2}+\frac{2 (d+e x)^{7/2}}{7 b}+\frac{(b d-a e)^4 \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{b^4}\\ &=\frac{2 (b d-a e)^3 \sqrt{d+e x}}{b^4}+\frac{2 (b d-a e)^2 (d+e x)^{3/2}}{3 b^3}+\frac{2 (b d-a e) (d+e x)^{5/2}}{5 b^2}+\frac{2 (d+e x)^{7/2}}{7 b}+\frac{\left (2 (b d-a e)^4\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^4 e}\\ &=\frac{2 (b d-a e)^3 \sqrt{d+e x}}{b^4}+\frac{2 (b d-a e)^2 (d+e x)^{3/2}}{3 b^3}+\frac{2 (b d-a e) (d+e x)^{5/2}}{5 b^2}+\frac{2 (d+e x)^{7/2}}{7 b}-\frac{2 (b d-a e)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.231581, size = 132, normalized size = 0.96 \[ \frac{2 (b d-a e) \left (5 (b d-a e) \left (\sqrt{b} \sqrt{d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )+3 b^{5/2} (d+e x)^{5/2}\right )}{15 b^{9/2}}+\frac{2 (d+e x)^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(7/2))/(7*b) + (2*(b*d - a*e)*(3*b^(5/2)*(d + e*x)^(5/2) + 5*(b*d - a*e)*(Sqrt[b]*Sqrt[d + e*x]*(
4*b*d - 3*a*e + b*e*x) - 3*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])))/(15*b^(9/2))

________________________________________________________________________________________

Maple [B]  time = 0.009, size = 380, normalized size = 2.8 \begin{align*}{\frac{2}{7\,b} \left ( ex+d \right ) ^{{\frac{7}{2}}}}-{\frac{2\,ae}{5\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{2\,d}{5\,b} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{a}^{2}{e}^{2}}{3\,{b}^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{4\,ade}{3\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{2\,{d}^{2}}{3\,b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-2\,{\frac{{e}^{3}{a}^{3}\sqrt{ex+d}}{{b}^{4}}}+6\,{\frac{{a}^{2}d{e}^{2}\sqrt{ex+d}}{{b}^{3}}}-6\,{\frac{a{d}^{2}e\sqrt{ex+d}}{{b}^{2}}}+2\,{\frac{{d}^{3}\sqrt{ex+d}}{b}}+2\,{\frac{{a}^{4}{e}^{4}}{{b}^{4}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-8\,{\frac{d{e}^{3}{a}^{3}}{{b}^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+12\,{\frac{{a}^{2}{d}^{2}{e}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-8\,{\frac{a{d}^{3}e}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+2\,{\frac{{d}^{4}}{\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(e*x+d)^(7/2)/b-2/5/b^2*(e*x+d)^(5/2)*a*e+2/5/b*(e*x+d)^(5/2)*d+2/3/b^3*(e*x+d)^(3/2)*a^2*e^2-4/3/b^2*(e*x
+d)^(3/2)*a*d*e+2/3/b*(e*x+d)^(3/2)*d^2-2/b^4*a^3*e^3*(e*x+d)^(1/2)+6/b^3*a^2*d*e^2*(e*x+d)^(1/2)-6/b^2*a*d^2*
e*(e*x+d)^(1/2)+2/b*d^3*(e*x+d)^(1/2)+2/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^
4*e^4-8/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*d*e^3*a^3+12/b^2/((a*e-b*d)*b)^(1/
2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2*d^2*e^2-8/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*
e-b*d)*b)^(1/2))*a*d^3*e+2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*d^4

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.05427, size = 940, normalized size = 6.81 \begin{align*} \left [-\frac{105 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e + 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) - 2 \,{\left (15 \, b^{3} e^{3} x^{3} + 176 \, b^{3} d^{3} - 406 \, a b^{2} d^{2} e + 350 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 3 \,{\left (22 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} +{\left (122 \, b^{3} d^{2} e - 112 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{105 \, b^{4}}, -\frac{2 \,{\left (105 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (15 \, b^{3} e^{3} x^{3} + 176 \, b^{3} d^{3} - 406 \, a b^{2} d^{2} e + 350 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 3 \,{\left (22 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} +{\left (122 \, b^{3} d^{2} e - 112 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}\right )}}{105 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/105*(105*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e
+ 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(15*b^3*e^3*x^3 + 176*b^3*d^3 - 406*a*b^2*d^2*e + 350*
a^2*b*d*e^2 - 105*a^3*e^3 + 3*(22*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 + (122*b^3*d^2*e - 112*a*b^2*d*e^2 + 35*a^2*b*e
^3)*x)*sqrt(e*x + d))/b^4, -2/105*(105*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-(b*d - a*e)/b
)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (15*b^3*e^3*x^3 + 176*b^3*d^3 - 406*a*b^2*d^2*e
+ 350*a^2*b*d*e^2 - 105*a^3*e^3 + 3*(22*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 + (122*b^3*d^2*e - 112*a*b^2*d*e^2 + 35*a
^2*b*e^3)*x)*sqrt(e*x + d))/b^4]

________________________________________________________________________________________

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((b*x+a)*(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.27989, size = 356, normalized size = 2.58 \begin{align*} \frac{2 \,{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{4}} + \frac{2 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{6} + 21 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{6} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{6} d^{2} + 105 \, \sqrt{x e + d} b^{6} d^{3} - 21 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{5} e - 70 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{5} d e - 315 \, \sqrt{x e + d} a b^{5} d^{2} e + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{4} e^{2} + 315 \, \sqrt{x e + d} a^{2} b^{4} d e^{2} - 105 \, \sqrt{x e + d} a^{3} b^{3} e^{3}\right )}}{105 \, b^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

2*(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d +
a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4) + 2/105*(15*(x*e + d)^(7/2)*b^6 + 21*(x*e + d)^(5/2)*b^6*d + 35*(x*e + d)^
(3/2)*b^6*d^2 + 105*sqrt(x*e + d)*b^6*d^3 - 21*(x*e + d)^(5/2)*a*b^5*e - 70*(x*e + d)^(3/2)*a*b^5*d*e - 315*sq
rt(x*e + d)*a*b^5*d^2*e + 35*(x*e + d)^(3/2)*a^2*b^4*e^2 + 315*sqrt(x*e + d)*a^2*b^4*d*e^2 - 105*sqrt(x*e + d)
*a^3*b^3*e^3)/b^7