3.2085 \(\int \frac{(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac{35 e^3 \sqrt{d+e x}}{64 b^4 (a+b x)}-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac{35 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{9/2} \sqrt{b d-a e}}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4} \]

[Out]

(-35*e^3*Sqrt[d + e*x])/(64*b^4*(a + b*x)) - (35*e^2*(d + e*x)^(3/2))/(96*b^3*(a + b*x)^2) - (7*e*(d + e*x)^(5
/2))/(24*b^2*(a + b*x)^3) - (d + e*x)^(7/2)/(4*b*(a + b*x)^4) - (35*e^4*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b
*d - a*e]])/(64*b^(9/2)*Sqrt[b*d - a*e])

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Rubi [A]  time = 0.0731864, antiderivative size = 152, 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, 47, 63, 208} \[ -\frac{35 e^3 \sqrt{d+e x}}{64 b^4 (a+b x)}-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac{35 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{9/2} \sqrt{b d-a e}}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(-35*e^3*Sqrt[d + e*x])/(64*b^4*(a + b*x)) - (35*e^2*(d + e*x)^(3/2))/(96*b^3*(a + b*x)^2) - (7*e*(d + e*x)^(5
/2))/(24*b^2*(a + b*x)^3) - (d + e*x)^(7/2)/(4*b*(a + b*x)^4) - (35*e^4*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b
*d - a*e]])/(64*b^(9/2)*Sqrt[b*d - a*e])

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 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 )^3} \, dx &=\int \frac{(d+e x)^{7/2}}{(a+b x)^5} \, dx\\ &=-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4}+\frac{(7 e) \int \frac{(d+e x)^{5/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4}+\frac{\left (35 e^2\right ) \int \frac{(d+e x)^{3/2}}{(a+b x)^3} \, dx}{48 b^2}\\ &=-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4}+\frac{\left (35 e^3\right ) \int \frac{\sqrt{d+e x}}{(a+b x)^2} \, dx}{64 b^3}\\ &=-\frac{35 e^3 \sqrt{d+e x}}{64 b^4 (a+b x)}-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4}+\frac{\left (35 e^4\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{128 b^4}\\ &=-\frac{35 e^3 \sqrt{d+e x}}{64 b^4 (a+b x)}-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4}+\frac{\left (35 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{64 b^4}\\ &=-\frac{35 e^3 \sqrt{d+e x}}{64 b^4 (a+b x)}-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4}-\frac{35 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{9/2} \sqrt{b d-a e}}\\ \end{align*}

Mathematica [A]  time = 0.232604, size = 152, normalized size = 1. \[ -\frac{35 e^3 \sqrt{d+e x}}{64 b^4 (a+b x)}-\frac{35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}+\frac{35 e^4 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{64 b^{9/2} \sqrt{a e-b d}}-\frac{7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{7/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(-35*e^3*Sqrt[d + e*x])/(64*b^4*(a + b*x)) - (35*e^2*(d + e*x)^(3/2))/(96*b^3*(a + b*x)^2) - (7*e*(d + e*x)^(5
/2))/(24*b^2*(a + b*x)^3) - (d + e*x)^(7/2)/(4*b*(a + b*x)^4) + (35*e^4*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(
b*d) + a*e]])/(64*b^(9/2)*Sqrt[-(b*d) + a*e])

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Maple [B]  time = 0.014, size = 318, normalized size = 2.1 \begin{align*} -{\frac{93\,{e}^{4}}{64\, \left ( bex+ae \right ) ^{4}b} \left ( ex+d \right ) ^{{\frac{7}{2}}}}-{\frac{511\,{e}^{5}a}{192\, \left ( bex+ae \right ) ^{4}{b}^{2}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{511\,{e}^{4}d}{192\, \left ( bex+ae \right ) ^{4}b} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{385\,{e}^{6}{a}^{2}}{192\, \left ( bex+ae \right ) ^{4}{b}^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{385\,{e}^{5}ad}{96\, \left ( bex+ae \right ) ^{4}{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{385\,{e}^{4}{d}^{2}}{192\, \left ( bex+ae \right ) ^{4}b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{35\,{e}^{7}{a}^{3}}{64\, \left ( bex+ae \right ) ^{4}{b}^{4}}\sqrt{ex+d}}+{\frac{105\,{e}^{6}d{a}^{2}}{64\, \left ( bex+ae \right ) ^{4}{b}^{3}}\sqrt{ex+d}}-{\frac{105\,{e}^{5}a{d}^{2}}{64\, \left ( bex+ae \right ) ^{4}{b}^{2}}\sqrt{ex+d}}+{\frac{35\,{e}^{4}{d}^{3}}{64\, \left ( bex+ae \right ) ^{4}b}\sqrt{ex+d}}+{\frac{35\,{e}^{4}}{64\,{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}}}} \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)^3,x)

[Out]

-93/64*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(7/2)-511/192*e^5/(b*e*x+a*e)^4/b^2*(e*x+d)^(5/2)*a+511/192*e^4/(b*e*x+a*e)
^4/b*(e*x+d)^(5/2)*d-385/192*e^6/(b*e*x+a*e)^4/b^3*(e*x+d)^(3/2)*a^2+385/96*e^5/(b*e*x+a*e)^4/b^2*(e*x+d)^(3/2
)*a*d-385/192*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(3/2)*d^2-35/64*e^7/(b*e*x+a*e)^4/b^4*(e*x+d)^(1/2)*a^3+105/64*e^6/(
b*e*x+a*e)^4/b^3*(e*x+d)^(1/2)*d*a^2-105/64*e^5/(b*e*x+a*e)^4/b^2*(e*x+d)^(1/2)*a*d^2+35/64*e^4/(b*e*x+a*e)^4/
b*(e*x+d)^(1/2)*d^3+35/64*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))

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.10182, size = 1602, normalized size = 10.54 \begin{align*} \left [\frac{105 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt{b^{2} d - a b e} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) - 2 \,{\left (48 \, b^{5} d^{4} + 8 \, a b^{4} d^{3} e + 14 \, a^{2} b^{3} d^{2} e^{2} + 35 \, a^{3} b^{2} d e^{3} - 105 \, a^{4} b e^{4} + 279 \,{\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} +{\left (326 \, b^{5} d^{2} e^{2} + 185 \, a b^{4} d e^{3} - 511 \, a^{2} b^{3} e^{4}\right )} x^{2} +{\left (200 \, b^{5} d^{3} e + 52 \, a b^{4} d^{2} e^{2} + 133 \, a^{2} b^{3} d e^{3} - 385 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt{e x + d}}{384 \,{\left (a^{4} b^{6} d - a^{5} b^{5} e +{\left (b^{10} d - a b^{9} e\right )} x^{4} + 4 \,{\left (a b^{9} d - a^{2} b^{8} e\right )} x^{3} + 6 \,{\left (a^{2} b^{8} d - a^{3} b^{7} e\right )} x^{2} + 4 \,{\left (a^{3} b^{7} d - a^{4} b^{6} e\right )} x\right )}}, \frac{105 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt{-b^{2} d + a b e} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) -{\left (48 \, b^{5} d^{4} + 8 \, a b^{4} d^{3} e + 14 \, a^{2} b^{3} d^{2} e^{2} + 35 \, a^{3} b^{2} d e^{3} - 105 \, a^{4} b e^{4} + 279 \,{\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} +{\left (326 \, b^{5} d^{2} e^{2} + 185 \, a b^{4} d e^{3} - 511 \, a^{2} b^{3} e^{4}\right )} x^{2} +{\left (200 \, b^{5} d^{3} e + 52 \, a b^{4} d^{2} e^{2} + 133 \, a^{2} b^{3} d e^{3} - 385 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt{e x + d}}{192 \,{\left (a^{4} b^{6} d - a^{5} b^{5} e +{\left (b^{10} d - a b^{9} e\right )} x^{4} + 4 \,{\left (a b^{9} d - a^{2} b^{8} e\right )} x^{3} + 6 \,{\left (a^{2} b^{8} d - a^{3} b^{7} e\right )} x^{2} + 4 \,{\left (a^{3} b^{7} d - a^{4} b^{6} e\right )} x\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)^3,x, algorithm="fricas")

[Out]

[1/384*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*
log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(48*b^5*d^4 + 8*a*b^4*d^3*e + 1
4*a^2*b^3*d^2*e^2 + 35*a^3*b^2*d*e^3 - 105*a^4*b*e^4 + 279*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (326*b^5*d^2*e^2 + 18
5*a*b^4*d*e^3 - 511*a^2*b^3*e^4)*x^2 + (200*b^5*d^3*e + 52*a*b^4*d^2*e^2 + 133*a^2*b^3*d*e^3 - 385*a^3*b^2*e^4
)*x)*sqrt(e*x + d))/(a^4*b^6*d - a^5*b^5*e + (b^10*d - a*b^9*e)*x^4 + 4*(a*b^9*d - a^2*b^8*e)*x^3 + 6*(a^2*b^8
*d - a^3*b^7*e)*x^2 + 4*(a^3*b^7*d - a^4*b^6*e)*x), 1/192*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*
x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) -
(48*b^5*d^4 + 8*a*b^4*d^3*e + 14*a^2*b^3*d^2*e^2 + 35*a^3*b^2*d*e^3 - 105*a^4*b*e^4 + 279*(b^5*d*e^3 - a*b^4*
e^4)*x^3 + (326*b^5*d^2*e^2 + 185*a*b^4*d*e^3 - 511*a^2*b^3*e^4)*x^2 + (200*b^5*d^3*e + 52*a*b^4*d^2*e^2 + 133
*a^2*b^3*d*e^3 - 385*a^3*b^2*e^4)*x)*sqrt(e*x + d))/(a^4*b^6*d - a^5*b^5*e + (b^10*d - a*b^9*e)*x^4 + 4*(a*b^9
*d - a^2*b^8*e)*x^3 + 6*(a^2*b^8*d - a^3*b^7*e)*x^2 + 4*(a^3*b^7*d - a^4*b^6*e)*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((b*x+a)*(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17753, size = 323, normalized size = 2.12 \begin{align*} \frac{35 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{4}}{64 \, \sqrt{-b^{2} d + a b e} b^{4}} - \frac{279 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{3} e^{4} - 511 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} d e^{4} + 385 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{4} - 105 \, \sqrt{x e + d} b^{3} d^{3} e^{4} + 511 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{2} e^{5} - 770 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{5} + 315 \, \sqrt{x e + d} a b^{2} d^{2} e^{5} + 385 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{6} - 315 \, \sqrt{x e + d} a^{2} b d e^{6} + 105 \, \sqrt{x e + d} a^{3} e^{7}}{192 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4} b^{4}} \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)^3,x, algorithm="giac")

[Out]

35/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/(sqrt(-b^2*d + a*b*e)*b^4) - 1/192*(279*(x*e + d)^(7/2)
*b^3*e^4 - 511*(x*e + d)^(5/2)*b^3*d*e^4 + 385*(x*e + d)^(3/2)*b^3*d^2*e^4 - 105*sqrt(x*e + d)*b^3*d^3*e^4 + 5
11*(x*e + d)^(5/2)*a*b^2*e^5 - 770*(x*e + d)^(3/2)*a*b^2*d*e^5 + 315*sqrt(x*e + d)*a*b^2*d^2*e^5 + 385*(x*e +
d)^(3/2)*a^2*b*e^6 - 315*sqrt(x*e + d)*a^2*b*d*e^6 + 105*sqrt(x*e + d)*a^3*e^7)/(((x*e + d)*b - b*d + a*e)^4*b
^4)