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

Optimal. Leaf size=197 \[ -\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {5 e^3 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}}-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {768, 646, 47, 63, 208} \begin {gather*} -\frac {5 e^2 \sqrt {d+e x}}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^3 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \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)^(5/2),x]

[Out]

-(d + e*x)^(5/2)/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (5*e^2*Sqrt[d + e*x])/(8*b^3*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (5*e*(d + e*x)^(3/2))/(12*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e^3*(a + b*x)*ArcTanh[(Sq
rt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*Sqrt[b*d - a*e]*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 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)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{6 b}\\ &=-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (5 b e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{\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)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e^2 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{\left (a b+b^2 x\right )^2} \, dx}{8 b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e^3 \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^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 e^2 \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^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{5/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^3 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 132, normalized size = 0.67 \begin {gather*} \frac {\frac {15 e^3 (a+b x)^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )}{\sqrt {a e-b d}}-\sqrt {b} \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (d+4 e x)+b^2 \left (8 d^2+26 d e x+33 e^2 x^2\right )\right )}{24 b^{7/2} \left ((a+b x)^2\right )^{3/2}} \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)^(5/2),x]

[Out]

(-(Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 26*d*e*x + 33*e^2*x^2))) + (15*e^3*
(a + b*x)^3*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/Sqrt[-(b*d) + a*e])/(24*b^(7/2)*((a + b*x)^2)^
(3/2))

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IntegrateAlgebraic [A]  time = 36.53, size = 187, normalized size = 0.95 \begin {gather*} \frac {(-a e-b e x) \left (\frac {e^3 \sqrt {d+e x} \left (15 a^2 e^2+40 a b e (d+e x)-30 a b d e+15 b^2 d^2+33 b^2 (d+e x)^2-40 b^2 d (d+e x)\right )}{24 b^3 (a e+b (d+e x)-b d)^3}+\frac {5 e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{7/2} \sqrt {a e-b d}}\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)^(5/2),x]

[Out]

((-(a*e) - b*e*x)*((e^3*Sqrt[d + e*x]*(15*b^2*d^2 - 30*a*b*d*e + 15*a^2*e^2 - 40*b^2*d*(d + e*x) + 40*a*b*e*(d
+ e*x) + 33*b^2*(d + e*x)^2))/(24*b^3*(-(b*d) + a*e + b*(d + e*x))^3) + (5*e^3*ArcTan[(Sqrt[b]*Sqrt[-(b*d) +
a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(8*b^(7/2)*Sqrt[-(b*d) + a*e])))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [A]  time = 0.45, size = 563, normalized size = 2.86 \begin {gather*} \left [\frac {15 \, {\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 {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} d^{3} + 2 \, a b^{3} d^{2} e + 5 \, a^{2} b^{2} d e^{2} - 15 \, a^{3} b e^{3} + 33 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} d^{2} e + 7 \, a b^{3} d e^{2} - 20 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d - a^{4} b^{4} e + {\left (b^{8} d - a b^{7} e\right )} x^{3} + 3 \, {\left (a b^{7} d - a^{2} b^{6} e\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d - a^{3} b^{5} e\right )} x\right )}}, \frac {15 \, {\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 {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (8 \, b^{4} d^{3} + 2 \, a b^{3} d^{2} e + 5 \, a^{2} b^{2} d e^{2} - 15 \, a^{3} b e^{3} + 33 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} d^{2} e + 7 \, a b^{3} d e^{2} - 20 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d - a^{4} b^{4} e + {\left (b^{8} d - a b^{7} e\right )} x^{3} + 3 \, {\left (a b^{7} d - a^{2} b^{6} e\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d - a^{3} b^{5} e\right )} x\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)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(15*(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^2*d - a*b*e)*log((b*e*x + 2*b*d - a
*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(8*b^4*d^3 + 2*a*b^3*d^2*e + 5*a^2*b^2*d*e^2 - 15*a^3
*b*e^3 + 33*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(13*b^4*d^2*e + 7*a*b^3*d*e^2 - 20*a^2*b^2*e^3)*x)*sqrt(e*x + d))/
(a^3*b^5*d - a^4*b^4*e + (b^8*d - a*b^7*e)*x^3 + 3*(a*b^7*d - a^2*b^6*e)*x^2 + 3*(a^2*b^6*d - a^3*b^5*e)*x), 1
/24*(15*(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^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*
b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (8*b^4*d^3 + 2*a*b^3*d^2*e + 5*a^2*b^2*d*e^2 - 15*a^3*b*e^3 + 33*(b^4*d*e^
2 - a*b^3*e^3)*x^2 + 2*(13*b^4*d^2*e + 7*a*b^3*d*e^2 - 20*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d - a^4*b^4*
e + (b^8*d - a*b^7*e)*x^3 + 3*(a*b^7*d - a^2*b^6*e)*x^2 + 3*(a^2*b^6*d - a^3*b^5*e)*x)]

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giac [A]  time = 0.29, size = 213, normalized size = 1.08 \begin {gather*} \frac {5 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, \sqrt {-b^{2} d + a b e} b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {33 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 40 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{3} + 15 \, \sqrt {x e + d} b^{2} d^{2} e^{3} + 40 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{4} - 30 \, \sqrt {x e + d} a b d e^{4} + 15 \, \sqrt {x e + d} a^{2} e^{5}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{3} \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)^(5/2),x, algorithm="giac")

[Out]

5/8*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/(sqrt(-b^2*d + a*b*e)*b^3*sgn((x*e + d)*b*e - b*d*e + a*e
^2)) - 1/24*(33*(x*e + d)^(5/2)*b^2*e^3 - 40*(x*e + d)^(3/2)*b^2*d*e^3 + 15*sqrt(x*e + d)*b^2*d^2*e^3 + 40*(x*
e + d)^(3/2)*a*b*e^4 - 30*sqrt(x*e + d)*a*b*d*e^4 + 15*sqrt(x*e + d)*a^2*e^5)/(((x*e + d)*b - b*d + a*e)^3*b^3
*sgn((x*e + d)*b*e - b*d*e + a*e^2))

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maple [B]  time = 0.11, size = 316, normalized size = 1.60 \begin {gather*} -\frac {\left (-15 b^{3} e^{3} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-45 a \,b^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-45 a^{2} b \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-15 a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} e^{2}-30 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b d e +15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d^{2}+40 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a b e -40 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} b^{2} d +33 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} b^{2}\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^{3}} \end {gather*}

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)^(5/2),x)

[Out]

-1/24*(-15*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*x^3*b^3*e^3-45*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)
*b)*x^2*a*b^2*e^3+33*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^2-45*a^2*b*e^3*x*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(
1/2)*b)+40*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b*e-40*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^2*d-15*a^3*e^3*arcta
n((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-30*((a*e-b*d)*b)^(1/2)*(e*
x+d)^(1/2)*a*b*d*e+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d^2)*(b*x+a)^2/((a*e-b*d)*b)^(1/2)/b^3/((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 {5}{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)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^(5/2)/(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 = -0.01 \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 )}^{5/2}} \,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)^(5/2),x)

[Out]

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

[Out]

Timed out

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