3.5.58 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\) [458]

Optimal. Leaf size=80 \[ \frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3873, 3853, 3855, 4131, 3852, 8} \begin {gather*} \frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a b \tan (c+d x) \sec (c+d x)}{d}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*ArcTanh[Sin[c + d*x]])/d + ((3*a^2 + 2*b^2)*Tan[c + d*x])/(3*d) + (a*b*Sec[c + d*x]*Tan[c + d*x])/d + (b^
2*Sec[c + d*x]^2*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 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 ^2(c+d x) (a+b \sec (c+d x))^2 \, dx &=(2 a b) \int \sec ^3(c+d x) \, dx+\int \sec ^2(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+(a b) \int \sec (c+d x) \, dx+\frac {1}{3} \left (3 a^2+2 b^2\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (3 a^2+2 b^2\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 71, normalized size = 0.89 \begin {gather*} \frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 74, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {a^{2} \tan \left (d x +c \right )+2 b a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(74\)
default	\(\frac {a^{2} \tan \left (d x +c \right )+2 b a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(74\)
norman	\(\frac {\frac {4 \left (3 a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (a^{2}-b a +b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (a^{2}+b a +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(132\)
risch	\(-\frac {2 i \left (3 b a \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 b a \,{\mathrm e}^{i \left (d x +c \right )}-3 a^{2}-2 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 84, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{2} - 3 \, a b {\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 )} + 6 \, a^{2} \tan \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.36, size = 100, normalized size = 1.25 \begin {gather*} \frac {3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).
time = 0.46, size = 178, normalized size = 2.22 \begin {gather*} \frac {3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \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 )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{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 )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*
x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*
b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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Mupad [B]
time = 3.07, size = 141, normalized size = 1.76 \begin {gather*} \frac {2\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,a^2-2\,a\,b+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,a^2-\frac {4\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+2\,a\,b+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\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 + b/cos(c + d*x))^2/cos(c + d*x)^2,x)

[Out]

(2*a*b*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)^5*(2*a^2 - 2*a*b + 2*b^2) - tan(c/2 + (d*x)/2)^3*(4*
a^2 + (4*b^2)/3) + tan(c/2 + (d*x)/2)*(2*a*b + 2*a^2 + 2*b^2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/
2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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