3.2.72 \(\int \frac {(c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^{3/2}} \, dx\) [172]

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

[Out]

2*d^3*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2)-1/2*(c-d)^3*tan(f*x+e)/a/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2
*c^3*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2
)-1/4*(c-d)^3*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*tan(f*x+e)/f*2^(1/2)/a^(1/2)/(a-a*sec(f*x+e)
)^(1/2)/(a+a*sec(f*x+e))^(1/2)-(c-d)^2*(c+2*d)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*tan
(f*x+e)/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 324, normalized size of antiderivative = 1.00, 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 = {4025, 186, 65, 212, 44} \begin {gather*} \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}} \end {gather*}

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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 186

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 212

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

Rule 4025

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]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 29.47, size = 21121, normalized size = 65.19 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(956\) vs. \(2(280)=560\).
time = 1.72, size = 957, normalized size = 2.95

method	result	size
default	\(\text {Expression too large to display}\)	\(957\)

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,method=_RETURNVERBOSE)

[Out]

1/4/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(-1+cos(f*x+e))*(4*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*
sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*2^(1/2)*sin(f*x+e)*c^3+5*ln((si
n(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c
os(f*x+e)*sin(f*x+e)*c^3-3*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*sin(f*x+e)*c^2*d-9*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1
/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*sin(f*x+e)*c*d^2+7*ln((sin(f*x+e
)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+
e)*sin(f*x+e)*d^3+4*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*2^(1/2)*(-
2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c^3*sin(f*x+e)+5*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*
x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c^3*sin(f*x+e)-3*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f
*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c^2*d*sin(f*x+e)-9*ln((sin(f*x+
e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*d^2*s
in(f*x+e)+7*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*(-2*cos(f*x+e)/(cos(
f*x+e)+1))^(1/2)*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*c^3*cos(f*x+e)-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/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [A]
time = 34.81, size = 744, normalized size = 2.30 \begin {gather*} \left [-\frac {\sqrt {2} {\left (5 \, c^{3} - 3 \, c^{2} d - 9 \, c d^{2} + 7 \, d^{3} + {\left (5 \, c^{3} - 3 \, c^{2} d - 9 \, c d^{2} + 7 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (5 \, c^{3} - 3 \, c^{2} d - 9 \, c d^{2} + 7 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (c^{3} \cos \left (f x + e\right )^{2} + 2 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 4 \, {\left (4 \, d^{3} - {\left (c^{3} - 3 \, c^{2} d + 3 \, c d^{2} - 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}}, \frac {\sqrt {2} {\left (5 \, c^{3} - 3 \, c^{2} d - 9 \, c d^{2} + 7 \, d^{3} + {\left (5 \, c^{3} - 3 \, c^{2} d - 9 \, c d^{2} + 7 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (5 \, c^{3} - 3 \, c^{2} d - 9 \, c d^{2} + 7 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - 8 \, {\left (c^{3} \cos \left (f x + e\right )^{2} + 2 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + 2 \, {\left (4 \, d^{3} - {\left (c^{3} - 3 \, c^{2} d + 3 \, c d^{2} - 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}}\right ] \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^3}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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