3.2130 \(\int \frac {(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=250 \[ -\frac {(d+e x)^{7/2}}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) \sqrt {d+e x} (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {768, 646, 50, 63, 208} \[ -\frac {(d+e x)^{7/2}}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) \sqrt {d+e x} (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.03, size = 66, normalized size = 0.26 \[ \frac {2 e (a+b x) (d+e x)^{9/2} \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};-\frac {b (d+e x)}{a e-b d}\right )}{9 \sqrt {(a+b x)^2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 486, normalized size = 1.94 \[ \left [\frac {105 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {105 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

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)^(3/2),x, algorithm="fricas")

[Out]

[1/30*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*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*(6*b^3*e^3*x^3 - 15*b^3*
d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e -
84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^5*x + a*b^4), -1/15*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a
^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d -
a*e)/b)/(b*d - a*e)) - (6*b^3*e^3*x^3 - 15*b^3*d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b
^3*d*e^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e - 84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^5*x + a*b^
4)]

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giac [B]  time = 0.30, size = 359, normalized size = 1.44 \[ \frac {7 \, {\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{\sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {{\left (\sqrt {x e + d} b^{3} d^{3} e^{2} - 3 \, \sqrt {x e + d} a b^{2} d^{2} e^{3} + 3 \, \sqrt {x e + d} a^{2} b d e^{4} - \sqrt {x e + d} a^{3} e^{5}\right )} e^{\left (-1\right )}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{8} e^{6} + 10 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{8} d e^{6} + 45 \, \sqrt {x e + d} b^{8} d^{2} e^{6} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{7} e^{7} - 90 \, \sqrt {x e + d} a b^{7} d e^{7} + 45 \, \sqrt {x e + d} a^{2} b^{6} e^{8}\right )} e^{\left (-5\right )}}{15 \, b^{10} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \]

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)^(3/2),x, algorithm="giac")

[Out]

7*(b^3*d^3*e^2 - 3*a*b^2*d^2*e^3 + 3*a^2*b*d*e^4 - a^3*e^5)*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)) - (sqrt(x*e + d)*b^3*d^3*e^2 - 3*sqrt(x*e + d)
*a*b^2*d^2*e^3 + 3*sqrt(x*e + d)*a^2*b*d*e^4 - sqrt(x*e + d)*a^3*e^5)*e^(-1)/(((x*e + d)*b - b*d + a*e)*b^4*sg
n((x*e + d)*b*e - b*d*e + a*e^2)) + 2/15*(3*(x*e + d)^(5/2)*b^8*e^6 + 10*(x*e + d)^(3/2)*b^8*d*e^6 + 45*sqrt(x
*e + d)*b^8*d^2*e^6 - 10*(x*e + d)^(3/2)*a*b^7*e^7 - 90*sqrt(x*e + d)*a*b^7*d*e^7 + 45*sqrt(x*e + d)*a^2*b^6*e
^8)*e^(-5)/(b^10*sgn((x*e + d)*b*e - b*d*e + a*e^2))

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maple [B]  time = 0.08, size = 662, normalized size = 2.65 \[ \frac {\left (-105 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 )-315 a \,b^{3} d^{2} e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 b^{4} d^{3} e 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 )+315 a^{3} b d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-315 a^{2} b^{2} d^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 a \,b^{3} d^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+90 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} b \,e^{3} x -180 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a \,b^{2} d \,e^{2} x +90 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{3} d^{2} e x +105 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-225 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+135 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e -20 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a \,b^{2} e^{2} x -15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}+20 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} b^{3} d e x -20 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{2}+20 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d e +6 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} b^{3} e x +6 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e \right ) \left (b x +a \right )^{2}}{15 \sqrt {\left (a e -b d \right ) b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \]

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)^(3/2),x)

[Out]

1/15*(6*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*x*b^3*e+6*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*a*b^2*e-20*((a*e-b*d)*b)
^(1/2)*(e*x+d)^(3/2)*x*a*b^2*e^2+20*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x*b^3*d*e-105*a^3*b*e^4*x*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*a^2*b^2*d*e^3-315*arctan((e*x+
d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*a*b^3*d^2*e^2+105*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x*b^4*d^3*e-20
*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a^2*b*e^2+20*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a*b^2*d*e+90*((a*e-b*d)*b)^(
1/2)*(e*x+d)^(1/2)*x*a^2*b*e^3-180*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*b^2*d*e^2+90*((a*e-b*d)*b)^(1/2)*(e*x
+d)^(1/2)*x*b^3*d^2*e-105*a^4*e^4*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)*a^3*b*d*e^3-315*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^2*b^2*d^2*e^2+105*arctan((e*x+d)
^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^3*d^3*e+105*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^3*e^3-225*(e*x+d)^(1/2)*((a*
e-b*d)*b)^(1/2)*a^2*b*d*e^2+135*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a*b^2*d^2*e-15*(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)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{\frac {7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^(7/2)/(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 \[ \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{7/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2),x)

[Out]

int(((a + b*x)*(d + e*x)^(7/2))/(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 \[ \text {Timed out} \]

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)**(3/2),x)

[Out]

Timed out

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