3.2.66 \(\int \frac {(c+d \sec (e+f x))^3}{\sqrt {a+a \sec (e+f x)}} \, dx\) [166]

Optimal. Leaf size=258 \[ \frac {2 (3 c-d) d^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 d^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^3 (1-\sec (e+f x)) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {2 \sqrt {a} c^3 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} \sqrt {a} (c-d)^3 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4025, 186, 65, 212, 45} \begin {gather*} \frac {2 \sqrt {a} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 d^2 (3 c-d) \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}}-\frac {\sqrt {2} \sqrt {a} (c-d)^3 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {2 d^3 \tan (e+f x) (1-\sec (e+f x))}{3 f \sqrt {a \sec (e+f x)+a}}+\frac {2 d^3 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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}{\sqrt {a+a \sec (e+f x)}} \, 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)} \, 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 {(3 c-d) d^2}{a \sqrt {a-a x}}+\frac {c^3}{a x \sqrt {a-a x}}+\frac {d^3 x}{a \sqrt {a-a x}}-\frac {(c-d)^3}{a (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 (3 c-d) d^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {\left (a 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 (a (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 )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (a d^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {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 (3 c-d) d^2 \tan (e+f x)}{f \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 )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (2 (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 )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (a d^3 \tan (e+f x)\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt {a-a x}}-\frac {\sqrt {a-a x}}{a}\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 (3 c-d) d^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 d^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^3 (1-\sec (e+f x)) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {2 \sqrt {a} 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 {\sqrt {2} \sqrt {a} (c-d)^3 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{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 5 vs. order 3 in optimal.
time = 7.26, size = 787, normalized size = 3.05 \begin {gather*} \frac {2 \cos \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 c \left (c^2+3 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{3 \left (1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )^{3/2}}-\frac {4 c^2 (c+3 d) \sin ^3\left (\frac {1}{2} (e+f x)\right )}{3 \left (1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )^{3/2}}+\frac {4 c \left (c^2+3 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{3 \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}}+\frac {1}{3} c^3 \csc \left (\frac {1}{2} (e+f x)\right ) \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \left (\frac {4 \sin ^4\left (\frac {1}{2} (e+f x)\right )}{\left (1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {6 \sin ^2\left (\frac {1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}+\frac {3 \sqrt {2} \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}}\right )-\frac {(c-d)^3 \csc ^5\left (\frac {1}{2} (e+f x)\right ) \left (-12 \cos ^4\left (\frac {1}{2} (e+f x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{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 ^8\left (\frac {1}{2} (e+f x)\right )-12 \, _2F_1\left (2,\frac {7}{2};\frac {9}{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 ^8\left (\frac {1}{2} (e+f x)\right ) \left (4-7 \sin ^2\left (\frac {1}{2} (e+f x)\right )+3 \sin ^4\left (\frac {1}{2} (e+f x)\right )\right )+7 \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 )^3 \left (15-20 \sin ^2\left (\frac {1}{2} (e+f x)\right )+8 \sin ^4\left (\frac {1}{2} (e+f x)\right )\right ) \left (\left (3-7 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {-\frac {\sin ^2\left (\frac {1}{2} (e+f x)\right )}{1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}}-3 \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 ) \left (1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{63 \left (1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )^{7/2}}\right )}{f (d+c \cos (e+f x))^3 \sec ^{\frac {5}{2}}(e+f x) \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[(e + f*x)/2]*(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
]*((2*c*(c^2 + 3*d^2)*Sin[(e + f*x)/2])/(3*(1 - 2*Sin[(e + f*x)/2]^2)^(3/2)) - (4*c^2*(c + 3*d)*Sin[(e + f*x)/
2]^3)/(3*(1 - 2*Sin[(e + f*x)/2]^2)^(3/2)) + (4*c*(c^2 + 3*d^2)*Sin[(e + f*x)/2])/(3*Sqrt[1 - 2*Sin[(e + f*x)/
2]^2]) + (c^3*Csc[(e + f*x)/2]*Sqrt[1 - 2*Sin[(e + f*x)/2]^2]*((4*Sin[(e + f*x)/2]^4)/(1 - 2*Sin[(e + f*x)/2]^
2)^2 - (6*Sin[(e + f*x)/2]^2)/(1 - 2*Sin[(e + f*x)/2]^2) + (3*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]*Sin[(e
+ f*x)/2])/Sqrt[1 - 2*Sin[(e + f*x)/2]^2]))/3 - ((c - d)^3*Csc[(e + f*x)/2]^5*(-12*Cos[(e + f*x)/2]^4*Hypergeo
metricPFQ[{2, 2, 7/2}, {1, 9/2}, -(Sin[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))]*Sin[(e + f*x)/2]^8 - 12*Hyp
ergeometric2F1[2, 7/2, 9/2, -(Sin[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))]*Sin[(e + f*x)/2]^8*(4 - 7*Sin[(e
+ f*x)/2]^2 + 3*Sin[(e + f*x)/2]^4) + 7*Sqrt[-(Sin[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))]*(1 - 2*Sin[(e
+ f*x)/2]^2)^3*(15 - 20*Sin[(e + f*x)/2]^2 + 8*Sin[(e + f*x)/2]^4)*((3 - 7*Sin[(e + f*x)/2]^2)*Sqrt[-(Sin[(e +
f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[(e + f*x)/2]^2/(1 - 2*Sin[(e + f*x)/2]^2))]]*(1
- 2*Sin[(e + f*x)/2]^2))))/(63*(1 - 2*Sin[(e + f*x)/2]^2)^(7/2))))/(f*(d + c*Cos[e + f*x])^3*Sec[e + f*x]^(5/
2)*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(906\) vs. \(2(227)=454\).
time = 1.53, size = 907, normalized size = 3.52

method	result	size
default	\(\text {Expression too large to display}\)	\(907\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(3*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))^(3/2)*cos(f*x+e)*2^(1/2)*sin(f*x+e)*c^3+3*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))^(3/2)*2^(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))^(3/2)*cos(f*x+e)*sin(f*x+e)*c^3-9*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/s
in(f*x+e))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*cos(f*x+e)*sin(f*x+e)*c^2*d+9*ln((sin(f*x+e)*(-2*cos(f*x+e)/(c
os(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))^(3/2)*cos(f*x+e)*sin(f*x+e)*c*d^2
-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)
)^(3/2)*cos(f*x+e)*sin(f*x+e)*d^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*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)*c^3*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))^(3/2)*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))^(3/2)*c*d^2*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))^(3/2)
*d^3*sin(f*x+e)-36*cos(f*x+e)^2*c*d^2+4*cos(f*x+e)^2*d^3+36*cos(f*x+e)*c*d^2-8*cos(f*x+e)*d^3+4*d^3)/sin(f*x+e
)/cos(f*x+e)/a

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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Fricas [A]
time = 23.42, size = 662, normalized size = 2.57 \begin {gather*} \left [-\frac {3 \, \sqrt {2} {\left ({\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 6 \, {\left (c^{3} \cos \left (f x + e\right )^{2} + c^{3} \cos \left (f x + e\right )\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 (d^{3} + {\left (9 \, c d^{2} - 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 )}{6 \, {\left (a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )\right )}}, -\frac {6 \, {\left (c^{3} \cos \left (f x + e\right )^{2} + c^{3} \cos \left (f x + e\right )\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 (d^{3} + {\left (9 \, c d^{2} - 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 ) - \frac {3 \, \sqrt {2} {\left ({\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right )\right )} \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 )}{\sqrt {a}}}{3 \, {\left (a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )\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))^(1/2),x, algorithm="fricas")

[Out]

[-1/6*(3*sqrt(2)*((a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3)*cos(f*x + e)^2 + (a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*
d^3)*cos(f*x + e))*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*
sin(f*x + e) - 3*cos(f*x + e)^2 - 2*cos(f*x + e) + 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 6*(c^3*cos(f*x
+ e)^2 + c^3*cos(f*x + e))*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*(d^3 + (9*c*d^2 - d^3)*cos(f*x + e)
)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a*f*cos(f*x + e)^2 + a*f*cos(f*x + e)), -1/3*(6*(c^3*
cos(f*x + e)^2 + c^3*cos(f*x + e))*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*(d^3 + (9*c*d^2 - d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)
- 3*sqrt(2)*((a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3)*cos(f*x + e)^2 + (a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3)
*cos(f*x + e))*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))/sqr
t(a))/(a*f*cos(f*x + e)^2 + a*f*cos(f*x + e))]

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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}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}\, 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))**(1/2),x)

[Out]

Integral((c + d*sec(e + f*x))**3/sqrt(a*(sec(e + f*x) + 1)), 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))^(1/2),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}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,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))^(1/2),x)

[Out]

int((c + d/cos(e + f*x))^3/(a + a/cos(e + f*x))^(1/2), x)

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