3.477 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=179 \[ \frac{2 \left (28 a^2 b^2+3 a^4+4 b^4\right ) \tan (c+d x)}{15 d}+\frac{a b \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (3 a^2+4 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{15 d}+\frac{a b \left (6 a^2+29 b^2\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^4}{5 d}+\frac{a \tan (c+d x) (a+b \sec (c+d x))^3}{5 d} \]

[Out]

(a*b*(4*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (2*(3*a^4 + 28*a^2*b^2 + 4*b^4)*Tan[c + d*x])/(15*d) + (a*
b*(6*a^2 + 29*b^2)*Sec[c + d*x]*Tan[c + d*x])/(30*d) + ((3*a^2 + 4*b^2)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(
15*d) + (a*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(5*d) + ((a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.301871, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3835, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{2 \left (28 a^2 b^2+3 a^4+4 b^4\right ) \tan (c+d x)}{15 d}+\frac{a b \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (3 a^2+4 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{15 d}+\frac{a b \left (6 a^2+29 b^2\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^4}{5 d}+\frac{a \tan (c+d x) (a+b \sec (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^4,x]

[Out]

(a*b*(4*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (2*(3*a^4 + 28*a^2*b^2 + 4*b^4)*Tan[c + d*x])/(15*d) + (a*
b*(6*a^2 + 29*b^2)*Sec[c + d*x]*Tan[c + d*x])/(30*d) + ((3*a^2 + 4*b^2)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(
15*d) + (a*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(5*d) + ((a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d)

Rule 3835

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

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

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

Rule 3770

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

Rule 3767

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 8

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

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{4}{5} \int \sec (c+d x) (b+a \sec (c+d x)) (a+b \sec (c+d x))^3 \, dx\\ &=\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (7 a b+\left (3 a^2+4 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{\left (3 a^2+4 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{15} \int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (27 a^2+8 b^2\right )+a \left (6 a^2+29 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{a b \left (6 a^2+29 b^2\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2+4 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{30} \int \sec (c+d x) \left (15 a b \left (4 a^2+3 b^2\right )+4 \left (3 a^4+28 a^2 b^2+4 b^4\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{a b \left (6 a^2+29 b^2\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2+4 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{2} \left (a b \left (4 a^2+3 b^2\right )\right ) \int \sec (c+d x) \, dx+\frac{1}{15} \left (2 \left (3 a^4+28 a^2 b^2+4 b^4\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a b \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a b \left (6 a^2+29 b^2\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2+4 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}-\frac{\left (2 \left (3 a^4+28 a^2 b^2+4 b^4\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a b \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 \left (3 a^4+28 a^2 b^2+4 b^4\right ) \tan (c+d x)}{15 d}+\frac{a b \left (6 a^2+29 b^2\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2+4 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.735435, size = 125, normalized size = 0.7 \[ \frac{15 a b \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (20 b^2 \left (3 a^2+b^2\right ) \tan ^2(c+d x)+15 a b \left (4 a^2+3 b^2\right ) \sec (c+d x)+30 \left (6 a^2 b^2+a^4+b^4\right )+30 a b^3 \sec ^3(c+d x)+6 b^4 \tan ^4(c+d x)\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^4,x]

[Out]

(15*a*b*(4*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(30*(a^4 + 6*a^2*b^2 + b^4) + 15*a*b*(4*a^2 + 3*b
^2)*Sec[c + d*x] + 30*a*b^3*Sec[c + d*x]^3 + 20*b^2*(3*a^2 + b^2)*Tan[c + d*x]^2 + 6*b^4*Tan[c + d*x]^4))/(30*
d)

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Maple [A]  time = 0.032, size = 225, normalized size = 1.3 \begin{align*}{\frac{{a}^{4}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{a{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,a{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,a{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{8\,{b}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^4,x)

[Out]

a^4*tan(d*x+c)/d+2/d*a^3*b*sec(d*x+c)*tan(d*x+c)+2/d*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+4/d*a^2*b^2*tan(d*x+c)+2/
d*a^2*b^2*tan(d*x+c)*sec(d*x+c)^2+1/d*a*b^3*tan(d*x+c)*sec(d*x+c)^3+3/2*a*b^3*sec(d*x+c)*tan(d*x+c)/d+3/2/d*a*
b^3*ln(sec(d*x+c)+tan(d*x+c))+8/15/d*b^4*tan(d*x+c)+1/5/d*b^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*b^4*tan(d*x+c)*se
c(d*x+c)^2

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Maxima [A]  time = 1.21169, size = 263, normalized size = 1.47 \begin{align*} \frac{120 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b^{2} + 4 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} b^{4} - 15 \, a b^{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 )} - 60 \, a^{3} 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 )} + 60 \, a^{4} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/60*(120*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b^2 + 4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*b^4 - 15*a*b^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)) - 60*a^3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1
) + log(sin(d*x + c) - 1)) + 60*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.7504, size = 443, normalized size = 2.47 \begin{align*} \frac{15 \,{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (30 \, a b^{3} \cos \left (d x + c\right ) + 2 \,{\left (15 \, a^{4} + 60 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, b^{4} + 15 \,{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (15 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/60*(15*(4*a^3*b + 3*a*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*a^3*b + 3*a*b^3)*cos(d*x + c)^5*log(
-sin(d*x + c) + 1) + 2*(30*a*b^3*cos(d*x + c) + 2*(15*a^4 + 60*a^2*b^2 + 8*b^4)*cos(d*x + c)^4 + 6*b^4 + 15*(4
*a^3*b + 3*a*b^3)*cos(d*x + c)^3 + 4*(15*a^2*b^2 + 2*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**4,x)

[Out]

Integral((a + b*sec(c + d*x))**4*sec(c + d*x)**2, x)

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Giac [B]  time = 1.39396, size = 622, normalized size = 3.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/30*(15*(4*a^3*b + 3*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*a^3*b + 3*a*b^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(30*a^4*tan(1/2*d*x + 1/2*c)^9 - 60*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*a^2*b^2*tan(1/2*d*x +
1/2*c)^9 - 75*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*b^4*tan(1/2*d*x + 1/2*c)^9 - 120*a^4*tan(1/2*d*x + 1/2*c)^7 +
120*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 480*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 30*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 40*
b^4*tan(1/2*d*x + 1/2*c)^7 + 180*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 116*b^4*tan
(1/2*d*x + 1/2*c)^5 - 120*a^4*tan(1/2*d*x + 1/2*c)^3 - 120*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 480*a^2*b^2*tan(1/2*
d*x + 1/2*c)^3 - 30*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 40*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*a^4*tan(1/2*d*x + 1/2*c)
+ 60*a^3*b*tan(1/2*d*x + 1/2*c) + 180*a^2*b^2*tan(1/2*d*x + 1/2*c) + 75*a*b^3*tan(1/2*d*x + 1/2*c) + 30*b^4*t
an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d