3.1.13 \(\int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\) [13]

Optimal. Leaf size=54 \[ \frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3873, 3852, 8, 4131, 3855} \begin {gather*} \frac {2 a^2 \tan (c+d x)}{d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*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 3855

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

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(219\) vs. \(2(54)=108\).
time = 0.65, size = 219, normalized size = 4.06 \begin {gather*} \frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {8 \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-6*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*Log[Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]] + (Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^(-2) - (Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^
(-2) + (8*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]))))/(16*d)

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Maple [A]
time = 0.04, size = 70, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} \tan \left (d x +c \right )+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(70\)
default	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} \tan \left (d x +c \right )+a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(70\)
norman	\(\frac {\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {3 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(95\)
risch	\(-\frac {i a^{2} \left ({\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+2*a^2*tan(d*x+c)+a^2*ln(sec(d*x+c)+tan(d*x+
c)))

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Maxima [A]
time = 0.28, size = 81, normalized size = 1.50 \begin {gather*} -\frac {a^{2} {\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 )} - 4 \, a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 8 \, a^{2} \tan \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*a^2*log(se
c(d*x + c) + tan(d*x + c)) - 8*a^2*tan(d*x + c))/d

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Fricas [A]
time = 2.51, size = 83, normalized size = 1.54 \begin {gather*} \frac {3 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*a^2*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(4*a^2*cos(d
*x + c) + a^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

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

[Out]

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

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Giac [A]
time = 0.46, size = 90, normalized size = 1.67 \begin {gather*} \frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} \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 )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*
x + 1/2*c)^3 - 5*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

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Mupad [B]
time = 1.18, size = 83, normalized size = 1.54 \begin {gather*} \frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\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))^2/cos(c + d*x),x)

[Out]

(3*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (3*a^2*tan(c/2 + (d*x)/2)^3 - 5*a^2*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d
*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1))

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