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

Optimal. Leaf size=468 \[ -\frac{\left (c^2-d^2\right ) \tan (e+f x)}{2 a^2 f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{\left (c^2-d^2\right ) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{3 (c-d)^2 \tan (e+f x)}{16 a^2 f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{(c-d)^2 \tan (e+f x)}{4 a^2 f (\sec (e+f x)+1)^2 \sqrt{a \sec (e+f x)+a}}-\frac{3 (c-d)^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{16 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

[Out]

-((c - d)^2*Tan[e + f*x])/(4*a^2*f*(1 + Sec[e + f*x])^2*Sqrt[a + a*Sec[e + f*x]]) - (3*(c - d)^2*Tan[e + f*x])
/(16*a^2*f*(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]) - ((c^2 - d^2)*Tan[e + f*x])/(2*a^2*f*(1 + Sec[e + f*x
])*Sqrt[a + a*Sec[e + f*x]]) + (2*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(a^(3/2)*f*Sqrt[
a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (Sqrt[2]*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a]
)]*Tan[e + f*x])/(a^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (3*(c - d)^2*ArcTanh[Sqrt[a -
a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(16*Sqrt[2]*a^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Se
c[e + f*x]]) - ((c^2 - d^2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(2*Sqrt[2]*a^(3/
2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.294868, antiderivative size = 468, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3940, 180, 63, 206, 51} \[ -\frac{\left (c^2-d^2\right ) \tan (e+f x)}{2 a^2 f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{\left (c^2-d^2\right ) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{3 (c-d)^2 \tan (e+f x)}{16 a^2 f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{(c-d)^2 \tan (e+f x)}{4 a^2 f (\sec (e+f x)+1)^2 \sqrt{a \sec (e+f x)+a}}-\frac{3 (c-d)^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{16 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c - d)^2*Tan[e + f*x])/(4*a^2*f*(1 + Sec[e + f*x])^2*Sqrt[a + a*Sec[e + f*x]]) - (3*(c - d)^2*Tan[e + f*x])
/(16*a^2*f*(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]) - ((c^2 - d^2)*Tan[e + f*x])/(2*a^2*f*(1 + Sec[e + f*x
])*Sqrt[a + a*Sec[e + f*x]]) + (2*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(a^(3/2)*f*Sqrt[
a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (Sqrt[2]*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a]
)]*Tan[e + f*x])/(a^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (3*(c - d)^2*ArcTanh[Sqrt[a -
a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(16*Sqrt[2]*a^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Se
c[e + f*x]]) - ((c^2 - d^2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(2*Sqrt[2]*a^(3/
2)*f*Sqrt[a - a*Sec[e + f*x]]*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])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rubi steps

\begin{align*} \int \frac{(c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^2}{x \sqrt{a-a x} (a+a x)^3} \, 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{c^2}{a^3 x \sqrt{a-a x}}-\frac{(c-d)^2}{a^3 (1+x)^3 \sqrt{a-a x}}+\frac{-c^2+d^2}{a^3 (1+x)^2 \sqrt{a-a x}}-\frac{c^2}{a^3 (1+x) \sqrt{a-a x}}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{\left (c^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (c^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left ((c-d)^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x)^3 \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (\left (c^2-d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x)^2 \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(c-d)^2 \tan (e+f x)}{4 a^2 f (1+\sec (e+f x))^2 \sqrt{a+a \sec (e+f x)}}-\frac{\left (c^2-d^2\right ) \tan (e+f x)}{2 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 c^2 \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 )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (2 c^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (3 (c-d)^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x)^2 \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{8 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (\left (c^2-d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{4 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(c-d)^2 \tan (e+f x)}{4 a^2 f (1+\sec (e+f x))^2 \sqrt{a+a \sec (e+f x)}}-\frac{3 (c-d)^2 \tan (e+f x)}{16 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}-\frac{\left (c^2-d^2\right ) \tan (e+f x)}{2 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{2 c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (3 (c-d)^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{32 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (\left (c^2-d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{2 a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(c-d)^2 \tan (e+f x)}{4 a^2 f (1+\sec (e+f x))^2 \sqrt{a+a \sec (e+f x)}}-\frac{3 (c-d)^2 \tan (e+f x)}{16 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}-\frac{\left (c^2-d^2\right ) \tan (e+f x)}{2 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{2 c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (c^2-d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{2 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (3 (c-d)^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{16 a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(c-d)^2 \tan (e+f x)}{4 a^2 f (1+\sec (e+f x))^2 \sqrt{a+a \sec (e+f x)}}-\frac{3 (c-d)^2 \tan (e+f x)}{16 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}-\frac{\left (c^2-d^2\right ) \tan (e+f x)}{2 a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{2 c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{3 (c-d)^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{16 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (c^2-d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{2 \sqrt{2} a^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 27.9488, size = 16259, normalized size = 34.74 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.208, size = 1133, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/f/a^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(32*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*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))*cos(f*x+e)^2*sin(f*x+e)*c^2+64*2^(1/2)*si
n(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*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^2*cos(f*x+e)+43*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1
/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)*c^2-6*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln
(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)*c*d-5*(-
2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x
+e))*cos(f*x+e)^2*sin(f*x+e)*d^2+32*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*2^(1/2)*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^2*sin(f*x+e)+86*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))
^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*c^2*cos(f*x+e)-12*sin(f
*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)
/sin(f*x+e))*c*d*cos(f*x+e)-10*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x
+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*d^2*cos(f*x+e)+43*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-
(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*c^2*sin(f*x+e)-6*(-2*cos(f*x+e)/(1+c
os(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*c*d*sin(f*x+
e)-5*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/
sin(f*x+e))*d^2*sin(f*x+e)-30*cos(f*x+e)^3*c^2+28*cos(f*x+e)^3*c*d+2*cos(f*x+e)^3*d^2+8*cos(f*x+e)^2*c^2-16*co
s(f*x+e)^2*c*d+8*cos(f*x+e)^2*d^2+22*c^2*cos(f*x+e)-12*cos(f*x+e)*c*d-10*cos(f*x+e)*d^2)/(1+cos(f*x+e))^2/sin(
f*x+e)

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

[Out]

Timed out

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Fricas [A]  time = 65.139, size = 1967, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*(sqrt(2)*((43*c^2 - 6*c*d - 5*d^2)*cos(f*x + e)^3 + 3*(43*c^2 - 6*c*d - 5*d^2)*cos(f*x + e)^2 + 43*c^2 -
6*c*d - 5*d^2 + 3*(43*c^2 - 6*c*d - 5*d^2)*cos(f*x + e))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(f*x + e
) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 3*a*cos(f*x + e)^2 + 2*a*cos(f*x + e) - a)/(cos(f*x + e)^2 +
2*cos(f*x + e) + 1)) - 64*(c^2*cos(f*x + e)^3 + 3*c^2*cos(f*x + e)^2 + 3*c^2*cos(f*x + e) + c^2)*sqrt(-a)*log(
(2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((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)) - 4*((15*c^2 - 14*c*d - d^2)*cos(f*x + e)^2 + (11*c^2 - 6*c*d - 5*d^2)*cos(f*x
+ e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3
*a^3*f*cos(f*x + e) + a^3*f), 1/32*(sqrt(2)*((43*c^2 - 6*c*d - 5*d^2)*cos(f*x + e)^3 + 3*(43*c^2 - 6*c*d - 5*d
^2)*cos(f*x + e)^2 + 43*c^2 - 6*c*d - 5*d^2 + 3*(43*c^2 - 6*c*d - 5*d^2)*cos(f*x + e))*sqrt(a)*arctan(sqrt(2)*
sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 64*(c^2*cos(f*x + e)^3 + 3*c^2*
cos(f*x + e)^2 + 3*c^2*cos(f*x + e) + c^2)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)
/(sqrt(a)*sin(f*x + e))) - 2*((15*c^2 - 14*c*d - d^2)*cos(f*x + e)^2 + (11*c^2 - 6*c*d - 5*d^2)*cos(f*x + e))*
sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f
*cos(f*x + e) + a^3*f)]

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

[Out]

Timed out

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

[Out]

Timed out