3.17.47 \(\int \frac {(A+B x) (d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=341 \[ -\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {(d+e x)^{5/2} (-7 a B e+3 A b e+4 b B d)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {5 e (a+b x) \sqrt {b d-a e} (-7 a B e+3 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (a+b x) \sqrt {d+e x} (-7 a B e+3 A b e+4 b B d)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (a+b x) (d+e x)^{3/2} (-7 a B e+3 A b e+4 b B d)}{12 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

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Rubi [A]  time = 0.30, antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {770, 78, 47, 50, 63, 208} \begin {gather*} -\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {(d+e x)^{5/2} (-7 a B e+3 A b e+4 b B d)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac {5 e (a+b x) (d+e x)^{3/2} (-7 a B e+3 A b e+4 b B d)}{12 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac {5 e (a+b x) \sqrt {d+e x} (-7 a B e+3 A b e+4 b B d)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (a+b x) \sqrt {b d-a e} (-7 a B e+3 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((4 b B d+3 A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{5/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(4 b B d+3 A b e-7 a B e) (d+e x)^{5/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e (4 b B d+3 A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{a b+b^2 x} \, dx}{8 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) (d+e x)^{3/2}}{12 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+3 A b e-7 a B e) (d+e x)^{5/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e \left (b^2 d-a b e\right ) (4 b B d+3 A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{8 b^4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) \sqrt {d+e x}}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) (d+e x)^{3/2}}{12 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+3 A b e-7 a B e) (d+e x)^{5/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e \left (b^2 d-a b e\right )^2 (4 b B d+3 A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 b^6 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) \sqrt {d+e x}}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) (d+e x)^{3/2}}{12 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+3 A b e-7 a B e) (d+e x)^{5/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 \left (b^2 d-a b e\right )^2 (4 b B d+3 A b e-7 a B e) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 b^6 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) \sqrt {d+e x}}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) (d+e x)^{3/2}}{12 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+3 A b e-7 a B e) (d+e x)^{5/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e \sqrt {b d-a e} (4 b B d+3 A b e-7 a B e) (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 111, normalized size = 0.33 \begin {gather*} \frac {(a+b x) (d+e x)^{7/2} \left (\frac {e (a+b x)^2 (-7 a B e+3 A b e+4 b B d) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+7 a B-7 A b\right )}{14 b \left ((a+b x)^2\right )^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((a + b*x)*(d + e*x)^(7/2)*(-7*A*b + 7*a*B + (e*(4*b*B*d + 3*A*b*e - 7*a*B*e)*(a + b*x)^2*Hypergeometric2F1[2,
7/2, 9/2, (b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^2))/(14*b*(b*d - a*e)*((a + b*x)^2)^(3/2))

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IntegrateAlgebraic [A]  time = 59.94, size = 372, normalized size = 1.09 \begin {gather*} \frac {(-a e-b e x) \left (\frac {5 \left (7 a^2 B e^3-3 a A b e^3-11 a b B d e^2+3 A b^2 d e^2+4 b^2 B d^2 e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{9/2} \sqrt {a e-b d}}-\frac {e \sqrt {d+e x} \left (-105 a^3 B e^3+45 a^2 A b e^3-175 a^2 b B e^2 (d+e x)+270 a^2 b B d e^2+75 a A b^2 e^2 (d+e x)-90 a A b^2 d e^2-225 a b^2 B d^2 e-56 a b^2 B e (d+e x)^2+275 a b^2 B d e (d+e x)+45 A b^3 d^2 e+24 A b^3 e (d+e x)^2-75 A b^3 d e (d+e x)+60 b^3 B d^3-100 b^3 B d^2 (d+e x)+8 b^3 B (d+e x)^3+32 b^3 B d (d+e x)^2\right )}{12 b^4 (a e+b (d+e x)-b d)^2}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^(5/2))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((-(a*e) - b*e*x)*(-1/12*(e*Sqrt[d + e*x]*(60*b^3*B*d^3 + 45*A*b^3*d^2*e - 225*a*b^2*B*d^2*e - 90*a*A*b^2*d*e^
2 + 270*a^2*b*B*d*e^2 + 45*a^2*A*b*e^3 - 105*a^3*B*e^3 - 100*b^3*B*d^2*(d + e*x) - 75*A*b^3*d*e*(d + e*x) + 27
5*a*b^2*B*d*e*(d + e*x) + 75*a*A*b^2*e^2*(d + e*x) - 175*a^2*b*B*e^2*(d + e*x) + 32*b^3*B*d*(d + e*x)^2 + 24*A
*b^3*e*(d + e*x)^2 - 56*a*b^2*B*e*(d + e*x)^2 + 8*b^3*B*(d + e*x)^3))/(b^4*(-(b*d) + a*e + b*(d + e*x))^2) + (
5*(4*b^2*B*d^2*e + 3*A*b^2*d*e^2 - 11*a*b*B*d*e^2 - 3*a*A*b*e^3 + 7*a^2*B*e^3)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a
*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*b^(9/2)*Sqrt[-(b*d) + a*e])))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [A]  time = 0.46, size = 680, normalized size = 1.99 \begin {gather*} \left [-\frac {15 \, {\left (4 \, B a^{2} b d e - {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\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 b^{3} e^{2} x^{3} - 6 \, {\left (B a b^{2} + A b^{3}\right )} d^{2} + 5 \, {\left (19 \, B a^{2} b - 3 \, A a b^{2}\right )} d e - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 8 \, {\left (7 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} - {\left (12 \, B b^{3} d^{2} - {\left (163 \, B a b^{2} - 27 \, A b^{3}\right )} d e + 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac {15 \, {\left (4 \, B a^{2} b d e - {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\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 b^{3} e^{2} x^{3} - 6 \, {\left (B a b^{2} + A b^{3}\right )} d^{2} + 5 \, {\left (19 \, B a^{2} b - 3 \, A a b^{2}\right )} d e - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 8 \, {\left (7 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} - {\left (12 \, B b^{3} d^{2} - {\left (163 \, B a b^{2} - 27 \, A b^{3}\right )} d e + 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/24*(15*(4*B*a^2*b*d*e - (7*B*a^3 - 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 + 2*(4*B
*a*b^2*d*e - (7*B*a^2*b - 3*A*a*b^2)*e^2)*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*b^3*e^2*x^3 - 6*(B*a*b^2 + A*b^3)*d^2 + 5*(19*B*a^2*b - 3*A*a*b^2)*d*
e - 15*(7*B*a^3 - 3*A*a^2*b)*e^2 + 8*(7*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 - (12*B*b^3*d^2 - (163*B*a*
b^2 - 27*A*b^3)*d*e + 25*(7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), -1/12
*(15*(4*B*a^2*b*d*e - (7*B*a^3 - 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2
*d*e - (7*B*a^2*b - 3*A*a*b^2)*e^2)*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*b^3*e^2*x^3 - 6*(B*a*b^2 + A*b^3)*d^2 + 5*(19*B*a^2*b - 3*A*a*b^2)*d*e - 15*(7*B*a^3 - 3*A*a^2*
b)*e^2 + 8*(7*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 - (12*B*b^3*d^2 - (163*B*a*b^2 - 27*A*b^3)*d*e + 25*(
7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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giac [A]  time = 0.37, size = 478, normalized size = 1.40 \begin {gather*} \frac {5 \, {\left (4 \, B b^{2} d^{2} e^{2} - 11 \, B a b d e^{3} + 3 \, A b^{2} d e^{3} + 7 \, B a^{2} e^{4} - 3 \, A a b e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{4 \, \sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {{\left (4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e^{2} - 4 \, \sqrt {x e + d} B b^{3} d^{3} e^{2} - 17 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} d e^{3} + 9 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} d e^{3} + 19 \, \sqrt {x e + d} B a b^{2} d^{2} e^{3} - 7 \, \sqrt {x e + d} A b^{3} d^{2} e^{3} + 13 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b e^{4} - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{2} e^{4} - 26 \, \sqrt {x e + d} B a^{2} b d e^{4} + 14 \, \sqrt {x e + d} A a b^{2} d e^{4} + 11 \, \sqrt {x e + d} B a^{3} e^{5} - 7 \, \sqrt {x e + d} A a^{2} b e^{5}\right )} e^{\left (-1\right )}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} B b^{6} e^{4} + 6 \, \sqrt {x e + d} B b^{6} d e^{4} - 9 \, \sqrt {x e + d} B a b^{5} e^{5} + 3 \, \sqrt {x e + d} A b^{6} e^{5}\right )} e^{\left (-3\right )}}{3 \, b^{9} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

5/4*(4*B*b^2*d^2*e^2 - 11*B*a*b*d*e^3 + 3*A*b^2*d*e^3 + 7*B*a^2*e^4 - 3*A*a*b*e^4)*arctan(sqrt(x*e + d)*b/sqrt
(-b^2*d + a*b*e))*e^(-1)/(sqrt(-b^2*d + a*b*e)*b^4*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1/4*(4*(x*e + d)^(3/2
)*B*b^3*d^2*e^2 - 4*sqrt(x*e + d)*B*b^3*d^3*e^2 - 17*(x*e + d)^(3/2)*B*a*b^2*d*e^3 + 9*(x*e + d)^(3/2)*A*b^3*d
*e^3 + 19*sqrt(x*e + d)*B*a*b^2*d^2*e^3 - 7*sqrt(x*e + d)*A*b^3*d^2*e^3 + 13*(x*e + d)^(3/2)*B*a^2*b*e^4 - 9*(
x*e + d)^(3/2)*A*a*b^2*e^4 - 26*sqrt(x*e + d)*B*a^2*b*d*e^4 + 14*sqrt(x*e + d)*A*a*b^2*d*e^4 + 11*sqrt(x*e + d
)*B*a^3*e^5 - 7*sqrt(x*e + d)*A*a^2*b*e^5)*e^(-1)/(((x*e + d)*b - b*d + a*e)^2*b^4*sgn((x*e + d)*b*e - b*d*e +
a*e^2)) + 2/3*((x*e + d)^(3/2)*B*b^6*e^4 + 6*sqrt(x*e + d)*B*b^6*d*e^4 - 9*sqrt(x*e + d)*B*a*b^5*e^5 + 3*sqrt
(x*e + d)*A*b^6*e^5)*e^(-3)/(b^9*sgn((x*e + d)*b*e - b*d*e + a*e^2))

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maple [B]  time = 0.08, size = 1150, normalized size = 3.37

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/12*(51*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*B*a*b^2*d*e-42*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*A*a*b^2*d*e^2+126*
((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*B*a^2*b*d*e^2-57*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*B*a*b^2*d^2*e+60*B*a^2*b^
2*d^2*e^2*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+21*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*A*b^3*d^2*e+45*((a*
e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*A*a^2*b*e^3+96*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*x*a*b^2*d*e^2+27*((a*e-b*d)*b
)^(1/2)*(e*x+d)^(3/2)*A*a*b^2*e^2-27*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*A*b^3*d*e+45*A*a^2*b^2*d*e^3*arctan((e*
x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-31*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*B*a^2*b*e^2-165*B*a^3*b*d*e^3*arctan((e
*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+105*B*a^4*e^4*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-45*A*a^3*b*e^4*ar
ctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-105*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*B*a^3*e^3-12*B*(e*x+d)^(3/2)*(
(a*e-b*d)*b)^(1/2)*b^3*d^2+12*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*b^3*d^3+48*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/
2)*x^2*b^3*d*e^2-330*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a^2*b^2*d*e^3+120*B*arctan((e*x+d)^(1/2)/
((a*e-b*d)*b)^(1/2)*b)*x*a*b^3*d^2*e^2+48*A*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*x*a*b^2*e^3+16*B*(e*x+d)^(3/2)*(
(a*e-b*d)*b)^(1/2)*x*a*b^2*e^2-72*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*x^2*a*b^2*e^3-144*B*(e*x+d)^(1/2)*((a*e-
b*d)*b)^(1/2)*x*a^2*b*e^3-165*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^2*a*b^3*d*e^3+90*A*arctan((e*x+d
)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a*b^3*d*e^3-90*A*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a^2*b^2*e^4+21
0*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a^3*b*e^4+8*B*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*x^2*b^3*e^2+
105*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^2*a^2*b^2*e^4+60*B*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2
)*b)*x^2*b^4*d^2*e^2-45*A*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^2*a*b^3*e^4+45*A*arctan((e*x+d)^(1/2)/
((a*e-b*d)*b)^(1/2)*b)*x^2*b^4*d*e^3+24*A*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*x^2*b^3*e^3)/e*(b*x+a)/((a*e-b*d)*
b)^(1/2)/b^4/((b*x+a)^2)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x + d)^(5/2)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{5/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(5/2))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(((A + B*x)*(d + e*x)^(5/2))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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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)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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