3.19.58 \(\int \frac {(a+b x) (d+e x)^{11/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=198 \[ -\frac {1155 e^4 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{13/2}}+\frac {1155 e^4 \sqrt {d+e x} (b d-a e)}{64 b^6}-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {385 e^4 (d+e x)^{3/2}}{64 b^5} \]

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Rubi [A]  time = 0.12, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \begin {gather*} -\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}+\frac {1155 e^4 \sqrt {d+e x} (b d-a e)}{64 b^6}-\frac {1155 e^4 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{13/2}}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {385 e^4 (d+e x)^{3/2}}{64 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1155*e^4*(b*d - a*e)*Sqrt[d + e*x])/(64*b^6) + (385*e^4*(d + e*x)^(3/2))/(64*b^5) - (231*e^3*(d + e*x)^(5/2))
/(64*b^4*(a + b*x)) - (33*e^2*(d + e*x)^(7/2))/(32*b^3*(a + b*x)^2) - (11*e*(d + e*x)^(9/2))/(24*b^2*(a + b*x)
^3) - (d + e*x)^(11/2)/(4*b*(a + b*x)^4) - (1155*e^4*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*
d - a*e]])/(64*b^(13/2))

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)^{11/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^{11/2}}{(a+b x)^5} \, dx\\ &=-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {(11 e) \int \frac {(d+e x)^{9/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {\left (33 e^2\right ) \int \frac {(d+e x)^{7/2}}{(a+b x)^3} \, dx}{16 b^2}\\ &=-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {\left (231 e^3\right ) \int \frac {(d+e x)^{5/2}}{(a+b x)^2} \, dx}{64 b^3}\\ &=-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {\left (1155 e^4\right ) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{128 b^4}\\ &=\frac {385 e^4 (d+e x)^{3/2}}{64 b^5}-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {\left (1155 e^4 (b d-a e)\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{128 b^5}\\ &=\frac {1155 e^4 (b d-a e) \sqrt {d+e x}}{64 b^6}+\frac {385 e^4 (d+e x)^{3/2}}{64 b^5}-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {\left (1155 e^4 (b d-a e)^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{128 b^6}\\ &=\frac {1155 e^4 (b d-a e) \sqrt {d+e x}}{64 b^6}+\frac {385 e^4 (d+e x)^{3/2}}{64 b^5}-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}+\frac {\left (1155 e^3 (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 )}{64 b^6}\\ &=\frac {1155 e^4 (b d-a e) \sqrt {d+e x}}{64 b^6}+\frac {385 e^4 (d+e x)^{3/2}}{64 b^5}-\frac {231 e^3 (d+e x)^{5/2}}{64 b^4 (a+b x)}-\frac {33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x)^2}-\frac {11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{11/2}}{4 b (a+b x)^4}-\frac {1155 e^4 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{13/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 52, normalized size = 0.26 \begin {gather*} \frac {2 e^4 (d+e x)^{13/2} \, _2F_1\left (5,\frac {13}{2};\frac {15}{2};-\frac {b (d+e x)}{a e-b d}\right )}{13 (a e-b d)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^4*(d + e*x)^(13/2)*Hypergeometric2F1[5, 13/2, 15/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(13*(-(b*d) + a*e)^
5)

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IntegrateAlgebraic [B]  time = 1.81, size = 436, normalized size = 2.20 \begin {gather*} \frac {1155 \left (-a^3 e^7+3 a^2 b d e^6-3 a b^2 d^2 e^5+b^3 d^3 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{64 b^{13/2} (a e-b d)^{3/2}}+\frac {e^4 \sqrt {d+e x} \left (-3465 a^5 e^5-12705 a^4 b e^4 (d+e x)+17325 a^4 b d e^4-34650 a^3 b^2 d^2 e^3-16863 a^3 b^2 e^3 (d+e x)^2+50820 a^3 b^2 d e^3 (d+e x)+34650 a^2 b^3 d^3 e^2-76230 a^2 b^3 d^2 e^2 (d+e x)-9207 a^2 b^3 e^2 (d+e x)^3+50589 a^2 b^3 d e^2 (d+e x)^2-17325 a b^4 d^4 e+50820 a b^4 d^3 e (d+e x)-50589 a b^4 d^2 e (d+e x)^2-1408 a b^4 e (d+e x)^4+18414 a b^4 d e (d+e x)^3+3465 b^5 d^5-12705 b^5 d^4 (d+e x)+16863 b^5 d^3 (d+e x)^2-9207 b^5 d^2 (d+e x)^3+128 b^5 (d+e x)^5+1408 b^5 d (d+e x)^4\right )}{192 b^6 (a e+b (d+e x)-b d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*Sqrt[d + e*x]*(3465*b^5*d^5 - 17325*a*b^4*d^4*e + 34650*a^2*b^3*d^3*e^2 - 34650*a^3*b^2*d^2*e^3 + 17325*a
^4*b*d*e^4 - 3465*a^5*e^5 - 12705*b^5*d^4*(d + e*x) + 50820*a*b^4*d^3*e*(d + e*x) - 76230*a^2*b^3*d^2*e^2*(d +
e*x) + 50820*a^3*b^2*d*e^3*(d + e*x) - 12705*a^4*b*e^4*(d + e*x) + 16863*b^5*d^3*(d + e*x)^2 - 50589*a*b^4*d^
2*e*(d + e*x)^2 + 50589*a^2*b^3*d*e^2*(d + e*x)^2 - 16863*a^3*b^2*e^3*(d + e*x)^2 - 9207*b^5*d^2*(d + e*x)^3 +
18414*a*b^4*d*e*(d + e*x)^3 - 9207*a^2*b^3*e^2*(d + e*x)^3 + 1408*b^5*d*(d + e*x)^4 - 1408*a*b^4*e*(d + e*x)^
4 + 128*b^5*(d + e*x)^5))/(192*b^6*(-(b*d) + a*e + b*(d + e*x))^4) + (1155*(b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*
a^2*b*d*e^6 - a^3*e^7)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(64*b^(13/2)*(-(b*d) +
a*e)^(3/2))

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fricas [B]  time = 0.46, size = 968, normalized size = 4.89 \begin {gather*} \left [-\frac {3465 \, {\left (a^{4} b d e^{4} - a^{5} e^{5} + {\left (b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 4 \, {\left (a b^{4} d e^{4} - a^{2} b^{3} e^{5}\right )} x^{3} + 6 \, {\left (a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 4 \, {\left (a^{3} b^{2} d e^{4} - a^{4} b e^{5}\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 (128 \, b^{5} e^{5} x^{5} - 48 \, b^{5} d^{5} - 88 \, a b^{4} d^{4} e - 198 \, a^{2} b^{3} d^{3} e^{2} - 693 \, a^{3} b^{2} d^{2} e^{3} + 4620 \, a^{4} b d e^{4} - 3465 \, a^{5} e^{5} + 128 \, {\left (16 \, b^{5} d e^{4} - 11 \, a b^{4} e^{5}\right )} x^{4} - {\left (2295 \, b^{5} d^{2} e^{3} - 12782 \, a b^{4} d e^{4} + 9207 \, a^{2} b^{3} e^{5}\right )} x^{3} - {\left (1030 \, b^{5} d^{3} e^{2} + 3795 \, a b^{4} d^{2} e^{3} - 22968 \, a^{2} b^{3} d e^{4} + 16863 \, a^{3} b^{2} e^{5}\right )} x^{2} - {\left (328 \, b^{5} d^{4} e + 748 \, a b^{4} d^{3} e^{2} + 2673 \, a^{2} b^{3} d^{2} e^{3} - 17094 \, a^{3} b^{2} d e^{4} + 12705 \, a^{4} b e^{5}\right )} x\right )} \sqrt {e x + d}}{384 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}}, -\frac {3465 \, {\left (a^{4} b d e^{4} - a^{5} e^{5} + {\left (b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 4 \, {\left (a b^{4} d e^{4} - a^{2} b^{3} e^{5}\right )} x^{3} + 6 \, {\left (a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 4 \, {\left (a^{3} b^{2} d e^{4} - a^{4} b e^{5}\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 (128 \, b^{5} e^{5} x^{5} - 48 \, b^{5} d^{5} - 88 \, a b^{4} d^{4} e - 198 \, a^{2} b^{3} d^{3} e^{2} - 693 \, a^{3} b^{2} d^{2} e^{3} + 4620 \, a^{4} b d e^{4} - 3465 \, a^{5} e^{5} + 128 \, {\left (16 \, b^{5} d e^{4} - 11 \, a b^{4} e^{5}\right )} x^{4} - {\left (2295 \, b^{5} d^{2} e^{3} - 12782 \, a b^{4} d e^{4} + 9207 \, a^{2} b^{3} e^{5}\right )} x^{3} - {\left (1030 \, b^{5} d^{3} e^{2} + 3795 \, a b^{4} d^{2} e^{3} - 22968 \, a^{2} b^{3} d e^{4} + 16863 \, a^{3} b^{2} e^{5}\right )} x^{2} - {\left (328 \, b^{5} d^{4} e + 748 \, a b^{4} d^{3} e^{2} + 2673 \, a^{2} b^{3} d^{2} e^{3} - 17094 \, a^{3} b^{2} d e^{4} + 12705 \, a^{4} b e^{5}\right )} x\right )} \sqrt {e x + d}}{192 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[-1/384*(3465*(a^4*b*d*e^4 - a^5*e^5 + (b^5*d*e^4 - a*b^4*e^5)*x^4 + 4*(a*b^4*d*e^4 - a^2*b^3*e^5)*x^3 + 6*(a^
2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 4*(a^3*b^2*d*e^4 - a^4*b*e^5)*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*(128*b^5*e^5*x^5 - 48*b^5*d^5 - 88*a*b^4*d^4*e - 198*
a^2*b^3*d^3*e^2 - 693*a^3*b^2*d^2*e^3 + 4620*a^4*b*d*e^4 - 3465*a^5*e^5 + 128*(16*b^5*d*e^4 - 11*a*b^4*e^5)*x^
4 - (2295*b^5*d^2*e^3 - 12782*a*b^4*d*e^4 + 9207*a^2*b^3*e^5)*x^3 - (1030*b^5*d^3*e^2 + 3795*a*b^4*d^2*e^3 - 2
2968*a^2*b^3*d*e^4 + 16863*a^3*b^2*e^5)*x^2 - (328*b^5*d^4*e + 748*a*b^4*d^3*e^2 + 2673*a^2*b^3*d^2*e^3 - 1709
4*a^3*b^2*d*e^4 + 12705*a^4*b*e^5)*x)*sqrt(e*x + d))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a
^4*b^6), -1/192*(3465*(a^4*b*d*e^4 - a^5*e^5 + (b^5*d*e^4 - a*b^4*e^5)*x^4 + 4*(a*b^4*d*e^4 - a^2*b^3*e^5)*x^3
+ 6*(a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 4*(a^3*b^2*d*e^4 - a^4*b*e^5)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*
x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (128*b^5*e^5*x^5 - 48*b^5*d^5 - 88*a*b^4*d^4*e - 198*a^2*b^3*d^3*
e^2 - 693*a^3*b^2*d^2*e^3 + 4620*a^4*b*d*e^4 - 3465*a^5*e^5 + 128*(16*b^5*d*e^4 - 11*a*b^4*e^5)*x^4 - (2295*b^
5*d^2*e^3 - 12782*a*b^4*d*e^4 + 9207*a^2*b^3*e^5)*x^3 - (1030*b^5*d^3*e^2 + 3795*a*b^4*d^2*e^3 - 22968*a^2*b^3
*d*e^4 + 16863*a^3*b^2*e^5)*x^2 - (328*b^5*d^4*e + 748*a*b^4*d^3*e^2 + 2673*a^2*b^3*d^2*e^3 - 17094*a^3*b^2*d*
e^4 + 12705*a^4*b*e^5)*x)*sqrt(e*x + d))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6)]

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giac [B]  time = 0.24, size = 476, normalized size = 2.40 \begin {gather*} \frac {1155 \, {\left (b^{2} d^{2} e^{4} - 2 \, a b d e^{5} + a^{2} e^{6}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, \sqrt {-b^{2} d + a b e} b^{6}} - \frac {2295 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{5} d^{2} e^{4} - 5855 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{5} d^{3} e^{4} + 5153 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{5} d^{4} e^{4} - 1545 \, \sqrt {x e + d} b^{5} d^{5} e^{4} - 4590 \, {\left (x e + d\right )}^{\frac {7}{2}} a b^{4} d e^{5} + 17565 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{4} d^{2} e^{5} - 20612 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{4} d^{3} e^{5} + 7725 \, \sqrt {x e + d} a b^{4} d^{4} e^{5} + 2295 \, {\left (x e + d\right )}^{\frac {7}{2}} a^{2} b^{3} e^{6} - 17565 \, {\left (x e + d\right )}^{\frac {5}{2}} a^{2} b^{3} d e^{6} + 30918 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{3} d^{2} e^{6} - 15450 \, \sqrt {x e + d} a^{2} b^{3} d^{3} e^{6} + 5855 \, {\left (x e + d\right )}^{\frac {5}{2}} a^{3} b^{2} e^{7} - 20612 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{3} b^{2} d e^{7} + 15450 \, \sqrt {x e + d} a^{3} b^{2} d^{2} e^{7} + 5153 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{4} b e^{8} - 7725 \, \sqrt {x e + d} a^{4} b d e^{8} + 1545 \, \sqrt {x e + d} a^{5} e^{9}}{192 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{4} b^{6}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{10} e^{4} + 15 \, \sqrt {x e + d} b^{10} d e^{4} - 15 \, \sqrt {x e + d} a b^{9} e^{5}\right )}}{3 \, b^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

1155/64*(b^2*d^2*e^4 - 2*a*b*d*e^5 + a^2*e^6)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*
e)*b^6) - 1/192*(2295*(x*e + d)^(7/2)*b^5*d^2*e^4 - 5855*(x*e + d)^(5/2)*b^5*d^3*e^4 + 5153*(x*e + d)^(3/2)*b^
5*d^4*e^4 - 1545*sqrt(x*e + d)*b^5*d^5*e^4 - 4590*(x*e + d)^(7/2)*a*b^4*d*e^5 + 17565*(x*e + d)^(5/2)*a*b^4*d^
2*e^5 - 20612*(x*e + d)^(3/2)*a*b^4*d^3*e^5 + 7725*sqrt(x*e + d)*a*b^4*d^4*e^5 + 2295*(x*e + d)^(7/2)*a^2*b^3*
e^6 - 17565*(x*e + d)^(5/2)*a^2*b^3*d*e^6 + 30918*(x*e + d)^(3/2)*a^2*b^3*d^2*e^6 - 15450*sqrt(x*e + d)*a^2*b^
3*d^3*e^6 + 5855*(x*e + d)^(5/2)*a^3*b^2*e^7 - 20612*(x*e + d)^(3/2)*a^3*b^2*d*e^7 + 15450*sqrt(x*e + d)*a^3*b
^2*d^2*e^7 + 5153*(x*e + d)^(3/2)*a^4*b*e^8 - 7725*sqrt(x*e + d)*a^4*b*d*e^8 + 1545*sqrt(x*e + d)*a^5*e^9)/(((
x*e + d)*b - b*d + a*e)^4*b^6) + 2/3*((x*e + d)^(3/2)*b^10*e^4 + 15*sqrt(x*e + d)*b^10*d*e^4 - 15*sqrt(x*e + d
)*a*b^9*e^5)/b^15

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maple [B]  time = 0.07, size = 701, normalized size = 3.54 \begin {gather*} -\frac {515 \sqrt {e x +d}\, a^{5} e^{9}}{64 \left (b e x +a e \right )^{4} b^{6}}+\frac {2575 \sqrt {e x +d}\, a^{4} d \,e^{8}}{64 \left (b e x +a e \right )^{4} b^{5}}-\frac {2575 \sqrt {e x +d}\, a^{3} d^{2} e^{7}}{32 \left (b e x +a e \right )^{4} b^{4}}+\frac {2575 \sqrt {e x +d}\, a^{2} d^{3} e^{6}}{32 \left (b e x +a e \right )^{4} b^{3}}-\frac {2575 \sqrt {e x +d}\, a \,d^{4} e^{5}}{64 \left (b e x +a e \right )^{4} b^{2}}+\frac {515 \sqrt {e x +d}\, d^{5} e^{4}}{64 \left (b e x +a e \right )^{4} b}-\frac {5153 \left (e x +d \right )^{\frac {3}{2}} a^{4} e^{8}}{192 \left (b e x +a e \right )^{4} b^{5}}+\frac {5153 \left (e x +d \right )^{\frac {3}{2}} a^{3} d \,e^{7}}{48 \left (b e x +a e \right )^{4} b^{4}}-\frac {5153 \left (e x +d \right )^{\frac {3}{2}} a^{2} d^{2} e^{6}}{32 \left (b e x +a e \right )^{4} b^{3}}+\frac {5153 \left (e x +d \right )^{\frac {3}{2}} a \,d^{3} e^{5}}{48 \left (b e x +a e \right )^{4} b^{2}}-\frac {5153 \left (e x +d \right )^{\frac {3}{2}} d^{4} e^{4}}{192 \left (b e x +a e \right )^{4} b}-\frac {5855 \left (e x +d \right )^{\frac {5}{2}} a^{3} e^{7}}{192 \left (b e x +a e \right )^{4} b^{4}}+\frac {5855 \left (e x +d \right )^{\frac {5}{2}} a^{2} d \,e^{6}}{64 \left (b e x +a e \right )^{4} b^{3}}-\frac {5855 \left (e x +d \right )^{\frac {5}{2}} a \,d^{2} e^{5}}{64 \left (b e x +a e \right )^{4} b^{2}}+\frac {5855 \left (e x +d \right )^{\frac {5}{2}} d^{3} e^{4}}{192 \left (b e x +a e \right )^{4} b}-\frac {765 \left (e x +d \right )^{\frac {7}{2}} a^{2} e^{6}}{64 \left (b e x +a e \right )^{4} b^{3}}+\frac {765 \left (e x +d \right )^{\frac {7}{2}} a d \,e^{5}}{32 \left (b e x +a e \right )^{4} b^{2}}-\frac {765 \left (e x +d \right )^{\frac {7}{2}} d^{2} e^{4}}{64 \left (b e x +a e \right )^{4} b}+\frac {1155 a^{2} e^{6} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{64 \sqrt {\left (a e -b d \right ) b}\, b^{6}}-\frac {1155 a d \,e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{32 \sqrt {\left (a e -b d \right ) b}\, b^{5}}+\frac {1155 d^{2} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{64 \sqrt {\left (a e -b d \right ) b}\, b^{4}}-\frac {10 \sqrt {e x +d}\, a \,e^{5}}{b^{6}}+\frac {10 \sqrt {e x +d}\, d \,e^{4}}{b^{5}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e^{4}}{3 b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2/3*e^4*(e*x+d)^(3/2)/b^5-10*e^5/b^6*a*(e*x+d)^(1/2)+10*e^4/b^5*(e*x+d)^(1/2)*d-765/64*e^6/b^3/(b*e*x+a*e)^4*(
e*x+d)^(7/2)*a^2+765/32*e^5/b^2/(b*e*x+a*e)^4*(e*x+d)^(7/2)*a*d-765/64*e^4/b/(b*e*x+a*e)^4*(e*x+d)^(7/2)*d^2-5
855/192*e^7/b^4/(b*e*x+a*e)^4*(e*x+d)^(5/2)*a^3+5855/64*e^6/b^3/(b*e*x+a*e)^4*(e*x+d)^(5/2)*a^2*d-5855/64*e^5/
b^2/(b*e*x+a*e)^4*(e*x+d)^(5/2)*a*d^2+5855/192*e^4/b/(b*e*x+a*e)^4*(e*x+d)^(5/2)*d^3-5153/192*e^8/b^5/(b*e*x+a
*e)^4*(e*x+d)^(3/2)*a^4+5153/48*e^7/b^4/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a^3*d-5153/32*e^6/b^3/(b*e*x+a*e)^4*(e*x+d
)^(3/2)*d^2*a^2+5153/48*e^5/b^2/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a*d^3-5153/192*e^4/b/(b*e*x+a*e)^4*(e*x+d)^(3/2)*d
^4-515/64*e^9/b^6/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^5+2575/64*e^8/b^5/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^4*d-2575/32*e^
7/b^4/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^3*d^2+2575/32*e^6/b^3/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^2*d^3-2575/64*e^5/b^2/
(b*e*x+a*e)^4*(e*x+d)^(1/2)*a*d^4+515/64*e^4/b/(b*e*x+a*e)^4*(e*x+d)^(1/2)*d^5+1155/64*e^6/b^6/((a*e-b*d)*b)^(
1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2-1155/32*e^5/b^5/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/
((a*e-b*d)*b)^(1/2)*b)*a*d+1155/64*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*d^2

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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)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^3,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 = 2.25, size = 535, normalized size = 2.70 \begin {gather*} \frac {2\,e^4\,{\left (d+e\,x\right )}^{3/2}}{3\,b^5}-\frac {{\left (d+e\,x\right )}^{7/2}\,\left (\frac {765\,a^2\,b^3\,e^6}{64}-\frac {765\,a\,b^4\,d\,e^5}{32}+\frac {765\,b^5\,d^2\,e^4}{64}\right )+\sqrt {d+e\,x}\,\left (\frac {515\,a^5\,e^9}{64}-\frac {2575\,a^4\,b\,d\,e^8}{64}+\frac {2575\,a^3\,b^2\,d^2\,e^7}{32}-\frac {2575\,a^2\,b^3\,d^3\,e^6}{32}+\frac {2575\,a\,b^4\,d^4\,e^5}{64}-\frac {515\,b^5\,d^5\,e^4}{64}\right )+{\left (d+e\,x\right )}^{5/2}\,\left (\frac {5855\,a^3\,b^2\,e^7}{192}-\frac {5855\,a^2\,b^3\,d\,e^6}{64}+\frac {5855\,a\,b^4\,d^2\,e^5}{64}-\frac {5855\,b^5\,d^3\,e^4}{192}\right )+{\left (d+e\,x\right )}^{3/2}\,\left (\frac {5153\,a^4\,b\,e^8}{192}-\frac {5153\,a^3\,b^2\,d\,e^7}{48}+\frac {5153\,a^2\,b^3\,d^2\,e^6}{32}-\frac {5153\,a\,b^4\,d^3\,e^5}{48}+\frac {5153\,b^5\,d^4\,e^4}{192}\right )}{b^{10}\,{\left (d+e\,x\right )}^4-\left (4\,b^{10}\,d-4\,a\,b^9\,e\right )\,{\left (d+e\,x\right )}^3+b^{10}\,d^4+{\left (d+e\,x\right )}^2\,\left (6\,a^2\,b^8\,e^2-12\,a\,b^9\,d\,e+6\,b^{10}\,d^2\right )-\left (d+e\,x\right )\,\left (-4\,a^3\,b^7\,e^3+12\,a^2\,b^8\,d\,e^2-12\,a\,b^9\,d^2\,e+4\,b^{10}\,d^3\right )+a^4\,b^6\,e^4-4\,a^3\,b^7\,d\,e^3+6\,a^2\,b^8\,d^2\,e^2-4\,a\,b^9\,d^3\,e}+\frac {2\,e^4\,\left (5\,b^5\,d-5\,a\,b^4\,e\right )\,\sqrt {d+e\,x}}{b^{10}}+\frac {1155\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^4\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^6-2\,a\,b\,d\,e^5+b^2\,d^2\,e^4}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{64\,b^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(d + e*x)^(11/2))/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

(2*e^4*(d + e*x)^(3/2))/(3*b^5) - ((d + e*x)^(7/2)*((765*a^2*b^3*e^6)/64 + (765*b^5*d^2*e^4)/64 - (765*a*b^4*d
*e^5)/32) + (d + e*x)^(1/2)*((515*a^5*e^9)/64 - (515*b^5*d^5*e^4)/64 + (2575*a*b^4*d^4*e^5)/64 - (2575*a^2*b^3
*d^3*e^6)/32 + (2575*a^3*b^2*d^2*e^7)/32 - (2575*a^4*b*d*e^8)/64) + (d + e*x)^(5/2)*((5855*a^3*b^2*e^7)/192 -
(5855*b^5*d^3*e^4)/192 + (5855*a*b^4*d^2*e^5)/64 - (5855*a^2*b^3*d*e^6)/64) + (d + e*x)^(3/2)*((5153*a^4*b*e^8
)/192 + (5153*b^5*d^4*e^4)/192 - (5153*a*b^4*d^3*e^5)/48 - (5153*a^3*b^2*d*e^7)/48 + (5153*a^2*b^3*d^2*e^6)/32
))/(b^10*(d + e*x)^4 - (4*b^10*d - 4*a*b^9*e)*(d + e*x)^3 + b^10*d^4 + (d + e*x)^2*(6*b^10*d^2 + 6*a^2*b^8*e^2
- 12*a*b^9*d*e) - (d + e*x)*(4*b^10*d^3 - 4*a^3*b^7*e^3 + 12*a^2*b^8*d*e^2 - 12*a*b^9*d^2*e) + a^4*b^6*e^4 -
4*a^3*b^7*d*e^3 + 6*a^2*b^8*d^2*e^2 - 4*a*b^9*d^3*e) + (2*e^4*(5*b^5*d - 5*a*b^4*e)*(d + e*x)^(1/2))/b^10 + (1
155*e^4*atan((b^(1/2)*e^4*(a*e - b*d)^(3/2)*(d + e*x)^(1/2))/(a^2*e^6 + b^2*d^2*e^4 - 2*a*b*d*e^5))*(a*e - b*d
)^(3/2))/(64*b^(13/2))

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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)**(11/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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