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

Optimal. Leaf size=324 \[ \frac{2 c^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{(c-d)^3 \tan (e+f x)}{2 a f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{(c-d)^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} \sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} (c-d)^2 (c+2 d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 d^3 \tan (e+f x)}{a f \sqrt{a \sec (e+f x)+a}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.236516, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{2 c^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{(c-d)^3 \tan (e+f x)}{2 a f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{(c-d)^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} \sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} (c-d)^2 (c+2 d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 d^3 \tan (e+f x)}{a f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^3*Tan[e + f*x])/(a*f*Sqrt[a + a*Sec[e + f*x]]) - ((c - d)^3*Tan[e + f*x])/(2*a*f*(1 + Sec[e + f*x])*Sqrt[
a + a*Sec[e + f*x]]) + (2*c^3*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(Sqrt[a]*f*Sqrt[a - a*Se
c[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - ((c - d)^3*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e +
f*x])/(2*Sqrt[2]*Sqrt[a]*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (Sqrt[2]*(c - d)^2*(c + 2*d)*
ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(Sqrt[a]*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))^3}{(a+a \sec (e+f x))^{3/2}} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^3}{x \sqrt{a-a x} (a+a x)^2} \, 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{d^3}{a^2 \sqrt{a-a x}}+\frac{c^3}{a^2 x \sqrt{a-a x}}-\frac{(c-d)^3}{a^2 (1+x)^2 \sqrt{a-a x}}-\frac{(c-d)^2 (c+2 d)}{a^2 (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{2 d^3 \tan (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{\left (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{\left ((c-d)^3 \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 )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left ((c-d)^2 (c+2 d) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{(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 d^3 \tan (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{(c-d)^3 \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 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 )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left ((c-d)^3 \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 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left (2 (c-d)^2 (c+2 d) \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 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 d^3 \tan (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{(c-d)^3 \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{2 c^3 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} (c-d)^2 (c+2 d) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\left ((c-d)^3 \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 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 d^3 \tan (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{(c-d)^3 \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}+\frac{2 c^3 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{(c-d)^3 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{2 \sqrt{2} \sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{\sqrt{2} (c-d)^2 (c+2 d) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right ) \tan (e+f x)}{\sqrt{a} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 6.5066, size = 856, normalized size = 2.64 \[ \frac{2 \cos ^3\left (\frac{1}{2} (e+f x)\right ) (c+d \sec (e+f x))^3 \sqrt{\frac{1}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}} \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )} \left (-\frac{2 \left (2 \sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right )+2 \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )} \sin \left (\frac{1}{2} (e+f x)\right )-\sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right )\right ) c^3}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}-\frac{4 (c-3 d) \sin \left (\frac{1}{2} (e+f x)\right ) c^2}{\sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}-\frac{3}{2} (c-d)^3 \tan ^{-1}\left (\frac{1-2 \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\frac{3}{2} (c-d)^3 \tan ^{-1}\left (\frac{2 \sin \left (\frac{1}{2} (e+f x)\right )+1}{\sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}\right )-\frac{(c-d)^2 (11 c+d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (5 \sqrt{-\frac{\sin ^2\left (\frac{1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}} \left (1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )^2 \left (3-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right ) \left (\sqrt{-\frac{\sin ^2\left (\frac{1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}-\tanh ^{-1}\left (\sqrt{-\frac{\sin ^2\left (\frac{1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right ) \csc ^4\left (\frac{1}{2} (e+f x)\right )+\frac{2 \cos ^2\left (\frac{1}{2} (e+f x)\right ) \text{Hypergeometric2F1}\left (2,\frac{5}{2},\frac{7}{2},-\frac{\sin ^2\left (\frac{1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}\right )}{10 \left (1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )^{3/2}}-\frac{(c-d)^3 \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}{1-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{(c-d)^3 \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}{\sin \left (\frac{1}{2} (e+f x)\right )+1}-\frac{(c-d)^3 \left (2 \sin \left (\frac{1}{2} (e+f x)\right )+1\right )}{4 \left (1-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}+\frac{(c-d)^3 \left (1-2 \sin \left (\frac{1}{2} (e+f x)\right )\right )}{4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+1\right ) \sqrt{1-2 \sin ^2\left (\frac{1}{2} (e+f x)\right )}}\right )}{f (d+c \cos (e+f x))^3 \sec ^{\frac{3}{2}}(e+f x) (a (\sec (e+f x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[(e + f*x)/2]^3*(c + d*Sec[e + f*x])^3*Sqrt[(1 - 2*Sin[(e + f*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[(e + f*x)/2]
^2]*((-3*(c - d)^3*ArcTan[(1 - 2*Sin[(e + f*x)/2])/Sqrt[1 - 2*Sin[(e + f*x)/2]^2]])/2 + (3*(c - d)^3*ArcTan[(1
+ 2*Sin[(e + f*x)/2])/Sqrt[1 - 2*Sin[(e + f*x)/2]^2]])/2 - (4*c^2*(c - 3*d)*Sin[(e + f*x)/2])/Sqrt[1 - 2*Sin[
(e + f*x)/2]^2] + ((c - d)^3*(1 - 2*Sin[(e + f*x)/2]))/(4*(1 + Sin[(e + f*x)/2])*Sqrt[1 - 2*Sin[(e + f*x)/2]^2
]) - ((c - d)^3*(1 + 2*Sin[(e + f*x)/2]))/(4*(1 - Sin[(e + f*x)/2])*Sqrt[1 - 2*Sin[(e + f*x)/2]^2]) - ((c - d)
^3*Sqrt[1 - 2*Sin[(e + f*x)/2]^2])/(1 - Sin[(e + f*x)/2]) + ((c - d)^3*Sqrt[1 - 2*Sin[(e + f*x)/2]^2])/(1 + Si
n[(e + f*x)/2]) - (2*c^3*(-(Sqrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]) + 2*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)
/2]]*Sin[(e + f*x)/2]^2 + 2*Sin[(e + f*x)/2]*Sqrt[1 - 2*Sin[(e + f*x)/2]^2]))/(1 - 2*Sin[(e + f*x)/2]^2) - ((c
- d)^2*(11*c + d)*Sin[(e + f*x)/2]*((2*Cos[(e + f*x)/2]^2*Hypergeometric2F1[2, 5/2, 7/2, -(Sin[(e + f*x)/2]^2
/(1 - 2*Sin[(e + f*x)/2]^2))]*Sin[(e + f*x)/2]^2)/(1 - 2*Sin[(e + f*x)/2]^2) + 5*Csc[(e + f*x)/2]^4*Sqrt[-(Sin
[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))]*(1 - 2*Sin[(e + f*x)/2]^2)^2*(3 - 2*Sin[(e + f*x)/2]^2)*(-ArcTanh
[Sqrt[-(Sin[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))]] + Sqrt[-(Sin[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2
))])))/(10*(1 - 2*Sin[(e + f*x)/2]^2)^(3/2))))/(f*(d + c*Cos[e + f*x])^3*Sec[e + f*x]^(3/2)*(a*(1 + Sec[e + f*
x]))^(3/2))

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Maple [B]  time = 0.268, size = 957, normalized size = 3. \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))^3/(a+a*sec(f*x+e))^(3/2),x)

[Out]

1/4/f/a^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(-1+cos(f*x+e))*(4*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*
x+e)*cos(f*x+e)*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^3+5*
(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*cos(f*x+e)*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^3-3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*cos(f*x+e)*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*d-9*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*s
in(f*x+e)*cos(f*x+e)*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^2+7*(
-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*cos(f*x+e)*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^3+4*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))*c^3*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))*c^3*sin(f*x+e)-3*(-2*cos(f*x+e)/(1+co
s(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*d*sin(f*x
+e)-9*(-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^2*sin(f*x+e)+7*(-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^3*sin(f*x+e)-2*cos(f*x+e)^2*c^3+6*cos(f*x+e)^2*c^2*d-6*cos(f*x+e)^2*c
*d^2+10*cos(f*x+e)^2*d^3+2*cos(f*x+e)*c^3-6*cos(f*x+e)*c^2*d+6*cos(f*x+e)*c*d^2-2*cos(f*x+e)*d^3-8*d^3)/sin(f*
x+e)^3

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

[Out]

Timed out

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Fricas [A]  time = 66.1323, size = 1740, normalized size = 5.37 \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))^3/(a+a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(2)*(5*c^3 - 3*c^2*d - 9*c*d^2 + 7*d^3 + (5*c^3 - 3*c^2*d - 9*c*d^2 + 7*d^3)*cos(f*x + e)^2 + 2*(5*
c^3 - 3*c^2*d - 9*c*d^2 + 7*d^3)*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)) + 8*(c^3*cos(f*x + e)^2 + 2*c^3*cos(f*x + e) + c^3)*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*s
qrt((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
*(4*d^3 - (c^3 - 3*c^2*d + 3*c*d^2 - 5*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)
)/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f), 1/4*(sqrt(2)*(5*c^3 - 3*c^2*d - 9*c*d^2 + 7*d^3 + (5*
c^3 - 3*c^2*d - 9*c*d^2 + 7*d^3)*cos(f*x + e)^2 + 2*(5*c^3 - 3*c^2*d - 9*c*d^2 + 7*d^3)*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))) - 8*(c^3*cos(f*x +
e)^2 + 2*c^3*cos(f*x + e) + c^3)*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*(4*d^3 - (c^3 - 3*c^2*d + 3*c*d^2 - 5*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x
+ e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d \sec{\left (e + f x \right )}\right )^{3}}{\left (a \left (\sec{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*sec(e + f*x))**3/(a*(sec(e + f*x) + 1))**(3/2), x)

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

[Out]

Timed out