3.160 \(\int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=336 \[ -\frac{2 \left (12 c^2 d+c^3+24 c d^2+12 d^3\right ) \tan (e+f x) \left (a^3-a^3 \sec (e+f x)\right )}{3 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^3 \left (12 c^2 d+3 c^3+12 c d^2+4 d^3\right ) \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{7/2} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a \sec (e+f x)+a} \sqrt{a-a \sec (e+f x)}}+\frac{2 a d \left (3 c^2+15 c d+13 d^2\right ) \tan (e+f x) (a-a \sec (e+f x))^2}{5 f \sqrt{a \sec (e+f x)+a}}-\frac{6 d^2 (c+2 d) \tan (e+f x) (a-a \sec (e+f x))^3}{7 f \sqrt{a \sec (e+f x)+a}}+\frac{2 d^3 \tan (e+f x) (a-a \sec (e+f x))^4}{9 a f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^3*(3*c^3 + 12*c^2*d + 12*c*d^2 + 4*d^3)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*c^3*ArcTa
nh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (2*
a*d*(3*c^2 + 15*c*d + 13*d^2)*(a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]) - (6*d^2*(c
+ 2*d)*(a - a*Sec[e + f*x])^3*Tan[e + f*x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) + (2*d^3*(a - a*Sec[e + f*x])^4*Tan
[e + f*x])/(9*a*f*Sqrt[a + a*Sec[e + f*x]]) - (2*(c^3 + 12*c^2*d + 24*c*d^2 + 12*d^3)*(a^3 - a^3*Sec[e + f*x])
*Tan[e + f*x])/(3*f*Sqrt[a + a*Sec[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.208889, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3940, 180, 63, 206} \[ -\frac{2 \left (12 c^2 d+c^3+24 c d^2+12 d^3\right ) \tan (e+f x) \left (a^3-a^3 \sec (e+f x)\right )}{3 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^3 \left (12 c^2 d+3 c^3+12 c d^2+4 d^3\right ) \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{7/2} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a \sec (e+f x)+a} \sqrt{a-a \sec (e+f x)}}+\frac{2 a d \left (3 c^2+15 c d+13 d^2\right ) \tan (e+f x) (a-a \sec (e+f x))^2}{5 f \sqrt{a \sec (e+f x)+a}}-\frac{6 d^2 (c+2 d) \tan (e+f x) (a-a \sec (e+f x))^3}{7 f \sqrt{a \sec (e+f x)+a}}+\frac{2 d^3 \tan (e+f x) (a-a \sec (e+f x))^4}{9 a f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x])^3,x]

[Out]

(2*a^3*(3*c^3 + 12*c^2*d + 12*c*d^2 + 4*d^3)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*c^3*ArcTa
nh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (2*
a*d*(3*c^2 + 15*c*d + 13*d^2)*(a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]) - (6*d^2*(c
+ 2*d)*(a - a*Sec[e + f*x])^3*Tan[e + f*x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) + (2*d^3*(a - a*Sec[e + f*x])^4*Tan
[e + f*x])/(9*a*f*Sqrt[a + a*Sec[e + f*x]]) - (2*(c^3 + 12*c^2*d + 24*c*d^2 + 12*d^3)*(a^3 - a^3*Sec[e + f*x])
*Tan[e + f*x])/(3*f*Sqrt[a + a*Sec[e + f*x]])

Rule 3940

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(a^2*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/(x*Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m - 1/2]

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x))^3 \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^2 (c+d x)^3}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \left (\frac{a^2 \left (3 c^3+12 c^2 d+12 c d^2+4 d^3\right )}{\sqrt{a-a x}}+\frac{a^2 c^3}{x \sqrt{a-a x}}-a \left (c^3+12 c^2 d+24 c d^2+12 d^3\right ) \sqrt{a-a x}+d \left (3 c^2+15 c d+13 d^2\right ) (a-a x)^{3/2}-\frac{3 d^2 (c+2 d) (a-a x)^{5/2}}{a}+\frac{d^3 (a-a x)^{7/2}}{a^2}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^3 \left (3 c^3+12 c^2 d+12 c d^2+4 d^3\right ) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a d \left (3 c^2+15 c d+13 d^2\right ) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}-\frac{6 d^2 (c+2 d) (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{2 d^3 (a-a \sec (e+f x))^4 \tan (e+f x)}{9 a f \sqrt{a+a \sec (e+f x)}}-\frac{2 \left (c^3+12 c^2 d+24 c d^2+12 d^3\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^4 c^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^3 \left (3 c^3+12 c^2 d+12 c d^2+4 d^3\right ) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a d \left (3 c^2+15 c d+13 d^2\right ) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}-\frac{6 d^2 (c+2 d) (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{2 d^3 (a-a \sec (e+f x))^4 \tan (e+f x)}{9 a f \sqrt{a+a \sec (e+f x)}}-\frac{2 \left (c^3+12 c^2 d+24 c d^2+12 d^3\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 a^3 c^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^3 \left (3 c^3+12 c^2 d+12 c d^2+4 d^3\right ) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^{7/2} c^3 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{2 a d \left (3 c^2+15 c d+13 d^2\right ) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}-\frac{6 d^2 (c+2 d) (a-a \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{2 d^3 (a-a \sec (e+f x))^4 \tan (e+f x)}{9 a f \sqrt{a+a \sec (e+f x)}}-\frac{2 \left (c^3+12 c^2 d+24 c d^2+12 d^3\right ) \left (a^3-a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.18927, size = 286, normalized size = 0.85 \[ \frac{a^2 \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt{a (\sec (e+f x)+1)} \left (2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\left (5292 c^2 d+630 c^3+7290 c d^2+2792 d^3\right ) \cos (e+f x)+4 \left (2898 c^2 d+840 c^3+2610 c d^2+803 d^3\right ) \cos (2 (e+f x))+1764 c^2 d \cos (3 (e+f x))+2709 c^2 d \cos (4 (e+f x))+8883 c^2 d+210 c^3 \cos (3 (e+f x))+840 c^3 \cos (4 (e+f x))+2520 c^3+2070 c d^2 \cos (3 (e+f x))+2070 c d^2 \cos (4 (e+f x))+8370 c d^2+584 d^3 \cos (3 (e+f x))+584 d^3 \cos (4 (e+f x))+2908 d^3\right )+2520 \sqrt{2} c^3 \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right ) \cos ^{\frac{9}{2}}(e+f x)\right )}{2520 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x])^3,x]

[Out]

(a^2*Sec[(e + f*x)/2]*Sec[e + f*x]^4*Sqrt[a*(1 + Sec[e + f*x])]*(2520*Sqrt[2]*c^3*ArcSin[Sqrt[2]*Sin[(e + f*x)
/2]]*Cos[e + f*x]^(9/2) + 2*(2520*c^3 + 8883*c^2*d + 8370*c*d^2 + 2908*d^3 + (630*c^3 + 5292*c^2*d + 7290*c*d^
2 + 2792*d^3)*Cos[e + f*x] + 4*(840*c^3 + 2898*c^2*d + 2610*c*d^2 + 803*d^3)*Cos[2*(e + f*x)] + 210*c^3*Cos[3*
(e + f*x)] + 1764*c^2*d*Cos[3*(e + f*x)] + 2070*c*d^2*Cos[3*(e + f*x)] + 584*d^3*Cos[3*(e + f*x)] + 840*c^3*Co
s[4*(e + f*x)] + 2709*c^2*d*Cos[4*(e + f*x)] + 2070*c*d^2*Cos[4*(e + f*x)] + 584*d^3*Cos[4*(e + f*x)])*Sin[(e
+ f*x)/2]))/(2520*f)

________________________________________________________________________________________

Maple [B]  time = 0.343, size = 677, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^3,x)

[Out]

-1/5040/f*a^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(315*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*(-2*cos(f*x+e)/(1+cos
(f*x+e)))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c^3+1260*sin(f
*x+e)*cos(f*x+e)^3*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+
e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c^3+1890*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2
)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c^3+1260*sin(f*x+e)*cos(f*x+
e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f
*x+e)/cos(f*x+e))*c^3+315*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+c
os(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c^3*sin(f*x+e)+26880*cos(f*x+e)^5*c^3+86688*cos(f*x+e)^5*c^2*d+66240*
cos(f*x+e)^5*c*d^2+18688*cos(f*x+e)^5*d^3-23520*cos(f*x+e)^4*c^3-58464*cos(f*x+e)^4*c^2*d-33120*cos(f*x+e)^4*c
*d^2-9344*cos(f*x+e)^4*d^3-3360*cos(f*x+e)^3*c^3-22176*cos(f*x+e)^3*c^2*d-15840*cos(f*x+e)^3*c*d^2-2336*cos(f*
x+e)^3*d^3-6048*cos(f*x+e)^2*c^2*d-12960*cos(f*x+e)^2*c*d^2-2848*cos(f*x+e)^2*d^3-4320*cos(f*x+e)*c*d^2-3040*c
os(f*x+e)*d^3-1120*d^3)/cos(f*x+e)^4/sin(f*x+e)

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.09471, size = 1513, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/315*(315*(a^2*c^3*cos(f*x + e)^5 + a^2*c^3*cos(f*x + e)^4)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sq
rt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*
(35*a^2*d^3 + (840*a^2*c^3 + 2709*a^2*c^2*d + 2070*a^2*c*d^2 + 584*a^2*d^3)*cos(f*x + e)^4 + (105*a^2*c^3 + 88
2*a^2*c^2*d + 1035*a^2*c*d^2 + 292*a^2*d^3)*cos(f*x + e)^3 + 3*(63*a^2*c^2*d + 180*a^2*c*d^2 + 73*a^2*d^3)*cos
(f*x + e)^2 + 5*(27*a^2*c*d^2 + 26*a^2*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)
)/(f*cos(f*x + e)^5 + f*cos(f*x + e)^4), -2/315*(315*(a^2*c^3*cos(f*x + e)^5 + a^2*c^3*cos(f*x + e)^4)*sqrt(a)
*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (35*a^2*d^3 + (840*a^2*
c^3 + 2709*a^2*c^2*d + 2070*a^2*c*d^2 + 584*a^2*d^3)*cos(f*x + e)^4 + (105*a^2*c^3 + 882*a^2*c^2*d + 1035*a^2*
c*d^2 + 292*a^2*d^3)*cos(f*x + e)^3 + 3*(63*a^2*c^2*d + 180*a^2*c*d^2 + 73*a^2*d^3)*cos(f*x + e)^2 + 5*(27*a^2
*c*d^2 + 26*a^2*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^5 + f
*cos(f*x + e)^4)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(5/2)*(c+d*sec(f*x+e))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e))^3,x, algorithm="giac")

[Out]

Timed out