3.20.13 \(\int \frac {(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=241 \[ -\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^3 (a+b x) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 241, 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 = {768, 646, 47, 50, 63, 208} \begin {gather*} -\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^3 (a+b x) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \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)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

-(d + e*x)^(7/2)/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) + (35*e^3*(a + b*x)*Sqrt[d + e*x])/(8*b^4*Sqrt[a^2 + 2*
a*b*x + b^2*x^2]) - (35*e^2*(d + e*x)^(3/2))/(24*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*e*(d + e*x)^(5/2))/(1
2*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*e^3*Sqrt[b*d - a*e]*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e
*x])/Sqrt[b*d - a*e]])/(8*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 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 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

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 )^{5/2}} \, dx &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{6 b}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (7 b e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^2 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{24 b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^3 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{16 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^3 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{16 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^2 \left (b^2 d-a b e\right ) \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 )}{8 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^3 \sqrt {b d-a e} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 68, normalized size = 0.28 \begin {gather*} \frac {2 e^3 (a+b x) (d+e x)^{9/2} \, _2F_1\left (4,\frac {9}{2};\frac {11}{2};-\frac {b (d+e x)}{a e-b d}\right )}{9 \sqrt {(a+b x)^2} (a e-b d)^4} \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)^(5/2),x]

[Out]

(2*e^3*(a + b*x)*(d + e*x)^(9/2)*Hypergeometric2F1[4, 9/2, 11/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(9*(-(b*d)
+ a*e)^4*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 48.23, size = 257, normalized size = 1.07 \begin {gather*} \frac {(-a e-b e x) \left (\frac {35 \left (b d e^3-a e^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{9/2} \sqrt {a e-b d}}-\frac {e^3 \sqrt {d+e x} \left (105 a^3 e^3+280 a^2 b e^2 (d+e x)-315 a^2 b d e^2+315 a b^2 d^2 e+231 a b^2 e (d+e x)^2-560 a b^2 d e (d+e x)-105 b^3 d^3+280 b^3 d^2 (d+e x)+48 b^3 (d+e x)^3-231 b^3 d (d+e x)^2\right )}{24 b^4 (a e+b (d+e x)-b d)^3}\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)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((-(a*e) - b*e*x)*(-1/24*(e^3*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 +
280*b^3*d^2*(d + e*x) - 560*a*b^2*d*e*(d + e*x) + 280*a^2*b*e^2*(d + e*x) - 231*b^3*d*(d + e*x)^2 + 231*a*b^2*
e*(d + e*x)^2 + 48*b^3*(d + e*x)^3))/(b^4*(-(b*d) + a*e + b*(d + e*x))^3) + (35*(b*d*e^3 - a*e^4)*ArcTan[(Sqrt
[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(8*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 = 498, normalized size = 2.07 \begin {gather*} \left [\frac {105 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + 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 (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \, {\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, -\frac {105 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + 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 (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \, {\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + 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*(48*b^3*e^3*x^3 - 8*b^3*d^3 - 14*a*b^2*d^2*e - 35*
a^2*b*d*e^2 + 105*a^3*e^3 - 3*(29*b^3*d*e^2 - 77*a*b^2*e^3)*x^2 - 2*(19*b^3*d^2*e + 49*a*b^2*d*e^2 - 140*a^2*b
*e^3)*x)*sqrt(e*x + d))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4), -1/24*(105*(b^3*e^3*x^3 + 3*a*b^2*e^3
*x^2 + 3*a^2*b*e^3*x + 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))
- (48*b^3*e^3*x^3 - 8*b^3*d^3 - 14*a*b^2*d^2*e - 35*a^2*b*d*e^2 + 105*a^3*e^3 - 3*(29*b^3*d*e^2 - 77*a*b^2*e^
3)*x^2 - 2*(19*b^3*d^2*e + 49*a*b^2*d*e^2 - 140*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^
5*x + a^3*b^4)]

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giac [A]  time = 0.35, size = 320, normalized size = 1.33 \begin {gather*} \frac {35 \, {\left (b d e^{3} - a e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \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 {2 \, \sqrt {x e + d} e^{3}}{b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {87 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} d e^{3} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{3} + 57 \, \sqrt {x e + d} b^{3} d^{3} e^{3} - 87 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{2} e^{4} + 272 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{4} - 171 \, \sqrt {x e + d} a b^{2} d^{2} e^{4} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{5} + 171 \, \sqrt {x e + d} a^{2} b d e^{5} - 57 \, \sqrt {x e + d} a^{3} e^{6}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{4} \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)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

35/8*(b*d*e^3 - a*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4*sgn((x*e + d)*b*
e - b*d*e + a*e^2)) + 2*sqrt(x*e + d)*e^3/(b^4*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1/24*(87*(x*e + d)^(5/2)*
b^3*d*e^3 - 136*(x*e + d)^(3/2)*b^3*d^2*e^3 + 57*sqrt(x*e + d)*b^3*d^3*e^3 - 87*(x*e + d)^(5/2)*a*b^2*e^4 + 27
2*(x*e + d)^(3/2)*a*b^2*d*e^4 - 171*sqrt(x*e + d)*a*b^2*d^2*e^4 - 136*(x*e + d)^(3/2)*a^2*b*e^5 + 171*sqrt(x*e
+ d)*a^2*b*d*e^5 - 57*sqrt(x*e + d)*a^3*e^6)/(((x*e + d)*b - b*d + a*e)^3*b^4*sgn((x*e + d)*b*e - b*d*e + a*e
^2))

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maple [B]  time = 0.07, size = 638, normalized size = 2.65 \begin {gather*} \frac {\left (-105 a \,b^{3} e^{4} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 b^{4} d \,e^{3} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-315 a^{2} b^{2} e^{4} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+315 a \,b^{3} d \,e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-315 a^{3} b \,e^{4} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+315 a^{2} b^{2} d \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-105 a^{4} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 a^{3} b d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+48 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{3} e^{3} x^{3}+144 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a \,b^{2} e^{3} x^{2}+144 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} b \,e^{3} x +105 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-171 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+171 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e -57 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}+136 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{2}-272 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d e +136 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2}+87 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e -87 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d \right ) \left (b x +a \right )^{2}}{24 \sqrt {\left (a e -b d \right ) b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{4}} \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)^(5/2),x)

[Out]

1/24*(-105*a*b^3*e^4*x^3*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+105*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1
/2)*b)*x^3*b^4*d*e^3+48*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^3*b^3*e^3-315*a^2*b^2*e^4*x^2*arctan((e*x+d)^(1/2)
/((a*e-b*d)*b)^(1/2)*b)+315*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^2*a*b^3*d*e^3+87*(e*x+d)^(5/2)*((a*e
-b*d)*b)^(1/2)*a*b^2*e-87*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*b^3*d+144*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*a*
b^2*e^3-315*a^3*b*e^4*x*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+315*a^2*b^2*d*e^3*x*arctan((e*x+d)^(1/2)/(
(a*e-b*d)*b)^(1/2)*b)+136*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a^2*b*e^2-272*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a*
b^2*d*e+136*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*b^3*d^2+144*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b*e^3*x-105*a^
4*e^4*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+105*a^3*b*d*e^3*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+
105*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^3*e^3-171*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^2*b*d*e^2+171*(e*x+d)^(1
/2)*((a*e-b*d)*b)^(1/2)*a*b^2*d^2*e-57*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*b^3*d^3)*(b*x+a)^2/((a*e-b*d)*b)^(1/2
)/b^4/((b*x+a)^2)^(5/2)

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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 {7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \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)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^(7/2)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), 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 )}^{7/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \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)^(5/2),x)

[Out]

int(((a + b*x)*(d + e*x)^(7/2))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), 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)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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