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

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

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Rubi [A]  time = 0.12, antiderivative size = 138, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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*}

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Mathematica [A]  time = 0.22, size = 132, normalized size = 0.96 \begin {gather*} \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} \end {gather*}

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))

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IntegrateAlgebraic [A]  time = 0.14, size = 190, normalized size = 1.38 \begin {gather*} \frac {2 \sqrt {d+e x} \left (-105 a^3 e^3+35 a^2 b e^2 (d+e x)+315 a^2 b d e^2-315 a b^2 d^2 e-21 a b^2 e (d+e x)^2-70 a b^2 d e (d+e x)+105 b^3 d^3+35 b^3 d^2 (d+e x)+15 b^3 (d+e x)^3+21 b^3 d (d+e x)^2\right )}{105 b^4}-\frac {2 (a e-b d)^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(105*b^3*d^3 - 315*a*b^2*d^2*e + 315*a^2*b*d*e^2 - 105*a^3*e^3 + 35*b^3*d^2*(d + e*x) - 70*a*
b^2*d*e*(d + e*x) + 35*a^2*b*e^2*(d + e*x) + 21*b^3*d*(d + e*x)^2 - 21*a*b^2*e*(d + e*x)^2 + 15*b^3*(d + e*x)^
3))/(105*b^4) - (2*(-(b*d) + a*e)^(7/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/b^(9/2
)

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fricas [A]  time = 0.44, size = 424, normalized size = 3.07 \begin {gather*} \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 {gather*}

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]

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giac [B]  time = 0.19, size = 264, normalized size = 1.91 \begin {gather*} \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 {gather*}

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

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maple [B]  time = 0.06, size = 380, normalized size = 2.75 \begin {gather*} \frac {2 a^{4} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{4}}-\frac {8 a^{3} d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {12 a^{2} d^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {8 a \,d^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 d^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}-\frac {2 \sqrt {e x +d}\, a^{3} e^{3}}{b^{4}}+\frac {6 \sqrt {e x +d}\, a^{2} d \,e^{2}}{b^{3}}-\frac {6 \sqrt {e x +d}\, a \,d^{2} e}{b^{2}}+\frac {2 \sqrt {e x +d}\, d^{3}}{b}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} a^{2} e^{2}}{3 b^{3}}-\frac {4 \left (e x +d \right )^{\frac {3}{2}} a d e}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} d^{2}}{3 b}-\frac {2 \left (e x +d \right )^{\frac {5}{2}} a e}{5 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} d}{5 b}+\frac {2 \left (e x +d \right )^{\frac {7}{2}}}{7 b} \end {gather*}

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 2.06, size = 165, normalized size = 1.20 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{7/2}}{7\,b}-\frac {2\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,b^2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{7/2}\,\sqrt {d+e\,x}}{a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4}\right )\,{\left (a\,e-b\,d\right )}^{7/2}}{b^{9/2}}+\frac {2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{3\,b^3}-\frac {2\,{\left (a\,e-b\,d\right )}^3\,\sqrt {d+e\,x}}{b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(d + e*x)^(7/2))/(7*b) - (2*(a*e - b*d)*(d + e*x)^(5/2))/(5*b^2) + (2*atan((b^(1/2)*(a*e - b*d)^(7/2)*(d +
e*x)^(1/2))/(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3))*(a*e - b*d)^(7/2))/b^(9/2
) + (2*(a*e - b*d)^2*(d + e*x)^(3/2))/(3*b^3) - (2*(a*e - b*d)^3*(d + e*x)^(1/2))/b^4

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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