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

Optimal. Leaf size=263 \[ \frac {2 (a+b x) (d+e x)^{5/2} (b d-a e)}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (b d-a e)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) \sqrt {d+e x} (b d-a e)^3}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{3/2} (b d-a e)^2}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 50, 63, 208} \begin {gather*} \frac {2 (a+b x) \sqrt {d+e x} (b d-a e)^3}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{3/2} (b d-a e)^2}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{5/2} (b d-a e)}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (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} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{7/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(d+e x)^{7/2}}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{5/2}}{a b+b^2 x} \, dx}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (b d-a e) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{a b+b^2 x} \, dx}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (b d-a e)^3 (a+b x) \sqrt {d+e x}}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (\left (b^2 d-a b e\right )^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{b^8 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (b d-a e)^3 (a+b x) \sqrt {d+e x}}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right )^4 \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^8 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (b d-a e)^3 (a+b x) \sqrt {d+e x}}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e)^2 (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (b d-a e) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{7/2}}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (b d-a e)^{7/2} (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 [A]  time = 0.25, size = 150, normalized size = 0.57 \begin {gather*} \frac {(a+b x) \left (\frac {14 (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^{7/2}}+2 (d+e x)^{7/2}\right )}{7 b \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(2*(d + e*x)^(7/2) + (14*(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^(7
/2))))/(7*b*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 46.27, size = 222, normalized size = 0.84 \begin {gather*} \frac {(-a e-b e x) \left (\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}}-\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}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(a*e) - b*e*x)*((-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)))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [A]  time = 0.43, size = 424, normalized size = 1.61 \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((e*x+d)^(7/2)/((b*x+a)^2)^(1/2),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 [A]  time = 0.27, size = 354, normalized size = 1.35 \begin {gather*} \frac {2 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\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} \mathrm {sgn}\left (b x + a\right ) + 21 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{6} d \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{6} d^{2} \mathrm {sgn}\left (b x + a\right ) + 105 \, \sqrt {x e + d} b^{6} d^{3} \mathrm {sgn}\left (b x + a\right ) - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{5} e \mathrm {sgn}\left (b x + a\right ) - 70 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{5} d e \mathrm {sgn}\left (b x + a\right ) - 315 \, \sqrt {x e + d} a b^{5} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 315 \, \sqrt {x e + d} a^{2} b^{4} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 105 \, \sqrt {x e + d} a^{3} b^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{105 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2*(b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x
+ a) + a^4*e^4*sgn(b*x + a))*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*sgn(b*x + a) + 21*(x*e + d)^(5/2)*b^6*d*sgn(b*x + a) + 35*(x*e + d)^(3/2)*b^6*d^2*sgn(b
*x + a) + 105*sqrt(x*e + d)*b^6*d^3*sgn(b*x + a) - 21*(x*e + d)^(5/2)*a*b^5*e*sgn(b*x + a) - 70*(x*e + d)^(3/2
)*a*b^5*d*e*sgn(b*x + a) - 315*sqrt(x*e + d)*a*b^5*d^2*e*sgn(b*x + a) + 35*(x*e + d)^(3/2)*a^2*b^4*e^2*sgn(b*x
+ a) + 315*sqrt(x*e + d)*a^2*b^4*d*e^2*sgn(b*x + a) - 105*sqrt(x*e + d)*a^3*b^3*e^3*sgn(b*x + a))/b^7

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maple [B]  time = 0.06, size = 462, normalized size = 1.76 \begin {gather*} \frac {2 \left (b x +a \right ) \left (105 a^{4} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-420 a^{3} b d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+630 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 )-420 a \,b^{3} d^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 b^{4} d^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-105 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}+315 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}-315 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e +105 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}+35 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{2}-70 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d e +35 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2}-21 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e +21 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d +15 \left (e x +d \right )^{\frac {7}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3}\right )}{105 \sqrt {\left (b x +a \right )^{2}}\, \sqrt {\left (a e -b d \right ) b}\, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x)

[Out]

2/105*(b*x+a)*(15*(e*x+d)^(7/2)*((a*e-b*d)*b)^(1/2)*b^3-21*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*a*b^2*e+21*(e*x+d
)^(5/2)*((a*e-b*d)*b)^(1/2)*b^3*d+35*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a^2*b*e^2-70*(e*x+d)^(3/2)*((a*e-b*d)*b
)^(1/2)*a*b^2*d*e+35*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*b^3*d^2+105*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)
*a^4*e^4-420*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a^3*b*d*e^3+630*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1
/2)*b)*a^2*b^2*d^2*e^2-420*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^3*d^3*e+105*arctan((e*x+d)^(1/2)/((
a*e-b*d)*b)^(1/2)*b)*b^4*d^4-105*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^3*e^3+315*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/
2)*a^2*b*d*e^2-315*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a*b^2*d^2*e+105*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*b^3*d^3
)/((b*x+a)^2)^(1/2)/b^4/((a*e-b*d)*b)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^{7/2}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(7/2)/((a + b*x)^2)^(1/2),x)

[Out]

int((d + e*x)^(7/2)/((a + b*x)^2)^(1/2), 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((e*x+d)**(7/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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