3.7 \(\int \sec ^7(a+b x) \, dx\)

Optimal. Leaf size=76 \[ \frac{5 \tanh ^{-1}(\sin (a+b x))}{16 b}+\frac{\tan (a+b x) \sec ^5(a+b x)}{6 b}+\frac{5 \tan (a+b x) \sec ^3(a+b x)}{24 b}+\frac{5 \tan (a+b x) \sec (a+b x)}{16 b} \]

[Out]

(5*ArcTanh[Sin[a + b*x]])/(16*b) + (5*Sec[a + b*x]*Tan[a + b*x])/(16*b) + (5*Sec[a + b*x]^3*Tan[a + b*x])/(24*
b) + (Sec[a + b*x]^5*Tan[a + b*x])/(6*b)

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Rubi [A]  time = 0.0386378, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3770} \[ \frac{5 \tanh ^{-1}(\sin (a+b x))}{16 b}+\frac{\tan (a+b x) \sec ^5(a+b x)}{6 b}+\frac{5 \tan (a+b x) \sec ^3(a+b x)}{24 b}+\frac{5 \tan (a+b x) \sec (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^7,x]

[Out]

(5*ArcTanh[Sin[a + b*x]])/(16*b) + (5*Sec[a + b*x]*Tan[a + b*x])/(16*b) + (5*Sec[a + b*x]^3*Tan[a + b*x])/(24*
b) + (Sec[a + b*x]^5*Tan[a + b*x])/(6*b)

Rule 3768

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 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 \sec ^7(a+b x) \, dx &=\frac{\sec ^5(a+b x) \tan (a+b x)}{6 b}+\frac{5}{6} \int \sec ^5(a+b x) \, dx\\ &=\frac{5 \sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac{\sec ^5(a+b x) \tan (a+b x)}{6 b}+\frac{5}{8} \int \sec ^3(a+b x) \, dx\\ &=\frac{5 \sec (a+b x) \tan (a+b x)}{16 b}+\frac{5 \sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac{\sec ^5(a+b x) \tan (a+b x)}{6 b}+\frac{5}{16} \int \sec (a+b x) \, dx\\ &=\frac{5 \tanh ^{-1}(\sin (a+b x))}{16 b}+\frac{5 \sec (a+b x) \tan (a+b x)}{16 b}+\frac{5 \sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac{\sec ^5(a+b x) \tan (a+b x)}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.166078, size = 52, normalized size = 0.68 \[ \frac{15 \tanh ^{-1}(\sin (a+b x))+\tan (a+b x) \sec (a+b x) \left (8 \sec ^4(a+b x)+10 \sec ^2(a+b x)+15\right )}{48 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^7,x]

[Out]

(15*ArcTanh[Sin[a + b*x]] + Sec[a + b*x]*(15 + 10*Sec[a + b*x]^2 + 8*Sec[a + b*x]^4)*Tan[a + b*x])/(48*b)

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Maple [A]  time = 0.046, size = 76, normalized size = 1. \begin{align*}{\frac{ \left ( \sec \left ( bx+a \right ) \right ) ^{5}\tan \left ( bx+a \right ) }{6\,b}}+{\frac{5\, \left ( \sec \left ( bx+a \right ) \right ) ^{3}\tan \left ( bx+a \right ) }{24\,b}}+{\frac{5\,\sec \left ( bx+a \right ) \tan \left ( bx+a \right ) }{16\,b}}+{\frac{5\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{16\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^7,x)

[Out]

1/6*sec(b*x+a)^5*tan(b*x+a)/b+5/24*sec(b*x+a)^3*tan(b*x+a)/b+5/16*sec(b*x+a)*tan(b*x+a)/b+5/16/b*ln(sec(b*x+a)
+tan(b*x+a))

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Maxima [A]  time = 1.07036, size = 123, normalized size = 1.62 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (b x + a\right )^{5} - 40 \, \sin \left (b x + a\right )^{3} + 33 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{96 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7,x, algorithm="maxima")

[Out]

-1/96*(2*(15*sin(b*x + a)^5 - 40*sin(b*x + a)^3 + 33*sin(b*x + a))/(sin(b*x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(
b*x + a)^2 - 1) - 15*log(sin(b*x + a) + 1) + 15*log(sin(b*x + a) - 1))/b

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Fricas [A]  time = 1.47555, size = 231, normalized size = 3.04 \begin{align*} \frac{15 \, \cos \left (b x + a\right )^{6} \log \left (\sin \left (b x + a\right ) + 1\right ) - 15 \, \cos \left (b x + a\right )^{6} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (15 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7,x, algorithm="fricas")

[Out]

1/96*(15*cos(b*x + a)^6*log(sin(b*x + a) + 1) - 15*cos(b*x + a)^6*log(-sin(b*x + a) + 1) + 2*(15*cos(b*x + a)^
4 + 10*cos(b*x + a)^2 + 8)*sin(b*x + a))/(b*cos(b*x + a)^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec ^{7}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**7,x)

[Out]

Integral(sec(a + b*x)**7, x)

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Giac [A]  time = 1.27607, size = 99, normalized size = 1.3 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (b x + a\right )^{5} - 40 \, \sin \left (b x + a\right )^{3} + 33 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{3}} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7,x, algorithm="giac")

[Out]

-1/96*(2*(15*sin(b*x + a)^5 - 40*sin(b*x + a)^3 + 33*sin(b*x + a))/(sin(b*x + a)^2 - 1)^3 - 15*log(abs(sin(b*x
+ a) + 1)) + 15*log(abs(sin(b*x + a) - 1)))/b