3.1.54 \(\int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [54]

Optimal. Leaf size=66 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {5 \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

arctanh(sin(d*x+c))/a^2/d-5/3*tan(d*x+c)/a^2/d/(1+sec(d*x+c))+1/3*tan(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]
time = 0.08, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3884, 4083, 3855, 3879} \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {5 \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(a^2*d) - (5*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + Tan[c + d*x]/(3*d*(a + a*Sec[c
+ d*x])^2)

Rule 3855

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

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3884

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*Cot[e + f*x]*((
a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(160\) vs. \(2(66)=132\).
time = 0.38, size = 160, normalized size = 2.42 \begin {gather*} -\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (6 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+8 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[(c + d*x)/2]*Sec[c + d*x]^2*(6*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sin[(d*x)/2] + 8*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + Cos[(
c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]
time = 0.07, size = 62, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(62\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(62\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+4\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\)	\(89\)
norman	\(\frac {-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 98, normalized size = 1.48 \begin {gather*} -\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(
d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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Fricas [A]
time = 3.12, size = 114, normalized size = 1.73 \begin {gather*} \frac {3 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 3*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*l
og(-sin(d*x + c) + 1) - 2*(4*cos(d*x + c) + 5)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^
2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 77, normalized size = 1.17 \begin {gather*} \frac {\frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - (a^4*tan(1/2*d*x +
1/2*c)^3 + 9*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 0.64, size = 43, normalized size = 0.65 \begin {gather*} -\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(9*tan(c/2 + (d*x)/2) - 12*atanh(tan(c/2 + (d*x)/2)) + tan(c/2 + (d*x)/2)^3)/(6*a^2*d)

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