3.33 \(\int (a+a \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=91 \[ \frac{5 a^4 \tan (c+d x)}{d}+\frac{6 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}+\frac{4 \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+a^4 x \]

[Out]

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

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Rubi [A]  time = 0.0898229, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3775, 3917, 3914, 3767, 8, 3770} \[ \frac{5 a^4 \tan (c+d x)}{d}+\frac{6 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}+\frac{4 \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+a^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^4,x]

[Out]

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

Rule 3775

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 3917

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

Rule 3914

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

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]

Rule 3770

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

Rubi steps

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

Mathematica [B]  time = 6.24585, size = 773, normalized size = 8.49 \[ \frac{1}{16} x \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4+\frac{5 \sin \left (\frac{d x}{2}\right ) \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4}{12 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{5 \sin \left (\frac{d x}{2}\right ) \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4}{12 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\left (13 \cos \left (\frac{c}{2}\right )-11 \sin \left (\frac{c}{2}\right )\right ) \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4}{192 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{\left (-11 \sin \left (\frac{c}{2}\right )-13 \cos \left (\frac{c}{2}\right )\right ) \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4}{192 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{\sin \left (\frac{d x}{2}\right ) \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4}{96 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{\sin \left (\frac{d x}{2}\right ) \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4}{96 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3}-\frac{3 \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4 \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{8 d}+\frac{3 \cos ^4(c+d x) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^4 \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^4,x]

[Out]

(x*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4)/16 - (3*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] -
Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4)/(8*d) + (3*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x
)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4)/(8*d) + (Cos[c + d*x]^4*Sec[c/2 + (d*x
)/2]^8*(a + a*Sec[c + d*x])^4*Sin[(d*x)/2])/(96*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/
2])^3) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(13*Cos[c/2] - 11*Sin[c/2]))/(192*d*(Cos[
c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (5*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^8*(a + a*
Sec[c + d*x])^4*Sin[(d*x)/2])/(12*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (Cos[c
+ d*x]^4*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*Sin[(d*x)/2])/(96*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x
)/2] + Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(-13*Cos[c/2] - 11
*Sin[c/2]))/(192*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (5*Cos[c + d*x]^4*Sec[
c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*Sin[(d*x)/2])/(12*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 = 0.036, size = 93, normalized size = 1. \begin{align*}{a}^{4}x+{\frac{{a}^{4}c}{d}}+6\,{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{20\,{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*x+1/d*a^4*c+6/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+20/3*a^4*tan(d*x+c)/d+2*a^4*sec(d*x+c)*tan(d*x+c)/d+1/3/d*a^
4*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.10326, size = 157, normalized size = 1.73 \begin{align*} a^{4} x + \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4}}{3 \, d} - \frac{a^{4}{\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 )}}{d} + \frac{4 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac{6 \, a^{4} \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.79029, size = 281, normalized size = 3.09 \begin{align*} \frac{3 \, a^{4} d x \cos \left (d x + c\right )^{3} + 9 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (20 \, a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^4*d*x*cos(d*x + c)^3 + 9*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 9*a^4*cos(d*x + c)^3*log(-sin(d*x
+ c) + 1) + (20*a^4*cos(d*x + c)^2 + 6*a^4*cos(d*x + c) + a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int 1\, dx + \int 4 \sec{\left (c + d x \right )}\, dx + \int 6 \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(Integral(1, x) + Integral(4*sec(c + d*x), x) + Integral(6*sec(c + d*x)**2, x) + Integral(4*sec(c + d*x)*
*3, x) + Integral(sec(c + d*x)**4, x))

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Giac [A]  time = 1.3023, size = 157, normalized size = 1.73 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{4} + 18 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 38 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, a^{4} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a^4 + 18*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(15*a^4*tan(1/2*d*x + 1/2*c)^5 - 38*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^3)/d