3.2075 \(\int \frac{(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=146 \[ \frac{35 e^2 \sqrt{d+e x} (b d-a e)}{4 b^4}-\frac{35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{9/2}}-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{35 e^2 (d+e x)^{3/2}}{12 b^3} \]

[Out]

(35*e^2*(b*d - a*e)*Sqrt[d + e*x])/(4*b^4) + (35*e^2*(d + e*x)^(3/2))/(12*b^3) - (7*e*(d + e*x)^(5/2))/(4*b^2*
(a + b*x)) - (d + e*x)^(7/2)/(2*b*(a + b*x)^2) - (35*e^2*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqr
t[b*d - a*e]])/(4*b^(9/2))

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Rubi [A]  time = 0.0759457, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {27, 47, 50, 63, 208} \[ \frac{35 e^2 \sqrt{d+e x} (b d-a e)}{4 b^4}-\frac{35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{9/2}}-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{35 e^2 (d+e x)^{3/2}}{12 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*e^2*(b*d - a*e)*Sqrt[d + e*x])/(4*b^4) + (35*e^2*(d + e*x)^(3/2))/(12*b^3) - (7*e*(d + e*x)^(5/2))/(4*b^2*
(a + b*x)) - (d + e*x)^(7/2)/(2*b*(a + b*x)^2) - (35*e^2*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqr
t[b*d - a*e]])/(4*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 47

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

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}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^{7/2}}{(a+b x)^3} \, dx\\ &=-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{(7 e) \int \frac{(d+e x)^{5/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{\left (35 e^2\right ) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{8 b^2}\\ &=\frac{35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{\left (35 e^2 (b d-a e)\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{8 b^3}\\ &=\frac{35 e^2 (b d-a e) \sqrt{d+e x}}{4 b^4}+\frac{35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{\left (35 e^2 (b d-a e)^2\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{8 b^4}\\ &=\frac{35 e^2 (b d-a e) \sqrt{d+e x}}{4 b^4}+\frac{35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac{\left (35 e (b d-a e)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 b^4}\\ &=\frac{35 e^2 (b d-a e) \sqrt{d+e x}}{4 b^4}+\frac{35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac{7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac{(d+e x)^{7/2}}{2 b (a+b x)^2}-\frac{35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0186399, size = 52, normalized size = 0.36 \[ \frac{2 e^2 (d+e x)^{9/2} \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};-\frac{b (d+e x)}{a e-b d}\right )}{9 (a e-b d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*(d + e*x)^(9/2)*Hypergeometric2F1[3, 9/2, 11/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(9*(-(b*d) + a*e)^3)

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Maple [B]  time = 0.016, size = 380, normalized size = 2.6 \begin{align*}{\frac{2\,{e}^{2}}{3\,{b}^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-6\,{\frac{{e}^{3}a\sqrt{ex+d}}{{b}^{4}}}+6\,{\frac{{e}^{2}d\sqrt{ex+d}}{{b}^{3}}}-{\frac{13\,{e}^{4}{a}^{2}}{4\,{b}^{3} \left ( bex+ae \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{13\,{e}^{3}ad}{2\,{b}^{2} \left ( bex+ae \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{13\,{e}^{2}{d}^{2}}{4\,b \left ( bex+ae \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{11\,{e}^{5}{a}^{3}}{4\,{b}^{4} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}+{\frac{33\,{e}^{4}d{a}^{2}}{4\,{b}^{3} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}-{\frac{33\,{e}^{3}a{d}^{2}}{4\,{b}^{2} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}+{\frac{11\,{e}^{2}{d}^{3}}{4\,b \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}+{\frac{35\,{e}^{4}{a}^{2}}{4\,{b}^{4}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}-{\frac{35\,{e}^{3}ad}{2\,{b}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{35\,{e}^{2}{d}^{2}}{4\,{b}^{2}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \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)^2,x)

[Out]

2/3*e^2*(e*x+d)^(3/2)/b^3-6*e^3/b^4*a*(e*x+d)^(1/2)+6*e^2/b^3*d*(e*x+d)^(1/2)-13/4*e^4/b^3/(b*e*x+a*e)^2*(e*x+
d)^(3/2)*a^2+13/2*e^3/b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)*a*d-13/4*e^2/b/(b*e*x+a*e)^2*(e*x+d)^(3/2)*d^2-11/4*e^5/
b^4/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a^3+33/4*e^4/b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d*a^2-33/4*e^3/b^2/(b*e*x+a*e)^2*
(e*x+d)^(1/2)*a*d^2+11/4*e^2/b/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d^3+35/4*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)
^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2-35/2*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))
*a*d+35/4*e^2/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*d^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((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.08774, size = 1094, normalized size = 7.49 \begin{align*} \left [-\frac{105 \,{\left (a^{2} b d e^{2} - a^{3} e^{3} +{\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\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 (8 \, b^{3} e^{3} x^{3} - 6 \, b^{3} d^{3} - 21 \, a b^{2} d^{2} e + 140 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 8 \,{\left (10 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} -{\left (39 \, b^{3} d^{2} e - 238 \, a b^{2} d e^{2} + 175 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{24 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac{105 \,{\left (a^{2} b d e^{2} - a^{3} e^{3} +{\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\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 (8 \, b^{3} e^{3} x^{3} - 6 \, b^{3} d^{3} - 21 \, a b^{2} d^{2} e + 140 \, a^{2} b d e^{2} - 105 \, a^{3} e^{3} + 8 \,{\left (10 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} -{\left (39 \, b^{3} d^{2} e - 238 \, a b^{2} d e^{2} + 175 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{12 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[-1/24*(105*(a^2*b*d*e^2 - a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2*b*e^3)*x)*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*(8*b^3*e^3*x^3 - 6*b^
3*d^3 - 21*a*b^2*d^2*e + 140*a^2*b*d*e^2 - 105*a^3*e^3 + 8*(10*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 - (39*b^3*d^2*e -
238*a*b^2*d*e^2 + 175*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), -1/12*(105*(a^2*b*d*e^2 -
a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2*b*e^3)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x +
d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^3*e^3*x^3 - 6*b^3*d^3 - 21*a*b^2*d^2*e + 140*a^2*b*d*e^2 - 105*
a^3*e^3 + 8*(10*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 - (39*b^3*d^2*e - 238*a*b^2*d*e^2 + 175*a^2*b*e^3)*x)*sqrt(e*x +
d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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

[Out]

Timed out

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Giac [B]  time = 1.25699, size = 358, normalized size = 2.45 \begin{align*} \frac{35 \,{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \, \sqrt{-b^{2} d + a b e} b^{4}} - \frac{13 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{2} - 11 \, \sqrt{x e + d} b^{3} d^{3} e^{2} - 26 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{3} + 33 \, \sqrt{x e + d} a b^{2} d^{2} e^{3} + 13 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{4} - 33 \, \sqrt{x e + d} a^{2} b d e^{4} + 11 \, \sqrt{x e + d} a^{3} e^{5}}{4 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{4}} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{6} e^{2} + 9 \, \sqrt{x e + d} b^{6} d e^{2} - 9 \, \sqrt{x e + d} a b^{5} e^{3}\right )}}{3 \, b^{9}} \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)^2,x, algorithm="giac")

[Out]

35/4*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*
b^4) - 1/4*(13*(x*e + d)^(3/2)*b^3*d^2*e^2 - 11*sqrt(x*e + d)*b^3*d^3*e^2 - 26*(x*e + d)^(3/2)*a*b^2*d*e^3 + 3
3*sqrt(x*e + d)*a*b^2*d^2*e^3 + 13*(x*e + d)^(3/2)*a^2*b*e^4 - 33*sqrt(x*e + d)*a^2*b*d*e^4 + 11*sqrt(x*e + d)
*a^3*e^5)/(((x*e + d)*b - b*d + a*e)^2*b^4) + 2/3*((x*e + d)^(3/2)*b^6*e^2 + 9*sqrt(x*e + d)*b^6*d*e^2 - 9*sqr
t(x*e + d)*a*b^5*e^3)/b^9