3.1.21 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \, dx\) [21]

Optimal. Leaf size=93 \[ \frac {15 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {15 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d} \]

[Out]

15/8*a^3*arctanh(sin(d*x+c))/d+4*a^3*tan(d*x+c)/d+15/8*a^3*sec(d*x+c)*tan(d*x+c)/d+1/4*a^3*sec(d*x+c)^3*tan(d*
x+c)/d+a^3*tan(d*x+c)^3/d

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Rubi [A]
time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3876, 3852, 8, 3853, 3855} \begin {gather*} \frac {a^3 \tan ^3(c+d x)}{d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {15 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {15 a^3 \tan (c+d x) \sec (c+d x)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (15*a^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^
3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*Tan[c + d*x]^3)/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+3 a^3 \sec ^4(c+d x)+a^3 \sec ^5(c+d x)\right ) \, dx\\ &=a^3 \int \sec ^2(c+d x) \, dx+a^3 \int \sec ^5(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \left (3 a^3\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{2} \left (3 a^3\right ) \int \sec (c+d x) \, dx-\frac {a^3 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (3 a^3\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {15 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d}+\frac {1}{8} \left (3 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac {15 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {15 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(877\) vs. \(2(93)=186\).
time = 6.43, size = 877, normalized size = 9.43 \begin {gather*} -\frac {15 \cos ^3(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3}{64 d}+\frac {15 \cos ^3(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3}{64 d}+\frac {\cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3}{128 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4}+\frac {\cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \sin \left (\frac {d x}{2}\right )}{16 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \left (19 \cos \left (\frac {c}{2}\right )-11 \sin \left (\frac {c}{2}\right )\right )}{128 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {3 \cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \sin \left (\frac {d x}{2}\right )}{8 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}-\frac {\cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3}{128 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4}+\frac {\cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \sin \left (\frac {d x}{2}\right )}{16 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \left (-19 \cos \left (\frac {c}{2}\right )-11 \sin \left (\frac {c}{2}\right )\right )}{128 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {3 \cos ^3(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \sin \left (\frac {d x}{2}\right )}{8 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Cos[c + d*x]^3*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/
(64*d) + (15*Cos[c + d*x]^3*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d
*x])^3)/(64*d) + (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/(128*d*(Cos[c/2 + (d*x)/2] - Sin
[c/2 + (d*x)/2])^4) + (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*Sin[(d*x)/2])/(16*d*(Cos[c/2
] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c
+ d*x])^3*(19*Cos[c/2] - 11*Sin[c/2]))/(128*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])
^2) + (3*Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*Sin[(d*x)/2])/(8*d*(Cos[c/2] - Sin[c/2])*(
Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/(128*
d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^4) + (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*S
in[(d*x)/2])/(16*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^3*Sec[c/
2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(-19*Cos[c/2] - 11*Sin[c/2]))/(128*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d
*x)/2] + Sin[c/2 + (d*x)/2])^2) + (3*Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*Sin[(d*x)/2])/
(8*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]
time = 0.07, size = 123, normalized size = 1.32

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \tan \left (d x +c \right )}{d}\)	\(123\)
default	\(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \tan \left (d x +c \right )}{d}\)	\(123\)
norman	\(\frac {\frac {49 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {73 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {55 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {15 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {15 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {15 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(133\)
risch	\(-\frac {i a^{3} \left (15 \,{\mathrm e}^{7 i \left (d x +c \right )}-8 \,{\mathrm e}^{6 i \left (d x +c \right )}+23 \,{\mathrm e}^{5 i \left (d x +c \right )}-72 \,{\mathrm e}^{4 i \left (d x +c \right )}-23 \,{\mathrm e}^{3 i \left (d x +c \right )}-88 \,{\mathrm e}^{2 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-24\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-3*a^3*(-2/3-1/3*sec(d*
x+c)^2)*tan(d*x+c)+3*a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a^3*tan(d*x+c))

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Maxima [A]
time = 0.29, size = 156, normalized size = 1.68 \begin {gather*} \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a^{3} \tan \left (d x + c\right )}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 -
2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*a^3*(2*sin(d*x + c)/(sin(d*x +
c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 16*a^3*tan(d*x + c))/d

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Fricas [A]
time = 2.14, size = 111, normalized size = 1.19 \begin {gather*} \frac {15 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, a^{3} \cos \left (d x + c\right )^{3} + 15 \, a^{3} \cos \left (d x + c\right )^{2} + 8 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(15*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*a^3*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*a^3*c
os(d*x + c)^3 + 15*a^3*cos(d*x + c)^2 + 8*a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(sec(c + d*x)**2, x) + Integral(3*sec(c + d*x)**3, x) + Integral(3*sec(c + d*x)**4, x) + Integra
l(sec(c + d*x)**5, x))

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Giac [A]
time = 0.49, size = 122, normalized size = 1.31 \begin {gather*} \frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 49 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(15*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*a^3*tan(1/2
*d*x + 1/2*c)^7 - 55*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*a^3*tan(1/2*d*x + 1/2*c)^3 - 49*a^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 4.05, size = 141, normalized size = 1.52 \begin {gather*} \frac {15\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {55\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {49\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*a^3*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((73*a^3*tan(c/2 + (d*x)/2)^3)/4 - (55*a^3*tan(c/2 + (d*x)/2)^5)/4
+ (15*a^3*tan(c/2 + (d*x)/2)^7)/4 - (49*a^3*tan(c/2 + (d*x)/2))/4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d
*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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