3.1.4 \(\int \sec ^4(a+b x) \, dx\) [4]

Optimal. Leaf size=26 \[ \frac {\tan (a+b x)}{b}+\frac {\tan ^3(a+b x)}{3 b} \]

[Out]

tan(b*x+a)/b+1/3*tan(b*x+a)^3/b

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Rubi [A]
time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3852} \begin {gather*} \frac {\tan ^3(a+b x)}{3 b}+\frac {\tan (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^4,x]

[Out]

Tan[a + b*x]/b + Tan[a + b*x]^3/(3*b)

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]

Rubi steps

\begin {align*} \int \sec ^4(a+b x) \, dx &=-\frac {\text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (a+b x)\right )}{b}\\ &=\frac {\tan (a+b x)}{b}+\frac {\tan ^3(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 23, normalized size = 0.88 \begin {gather*} \frac {\tan (a+b x)+\frac {1}{3} \tan ^3(a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^4,x]

[Out]

(Tan[a + b*x] + Tan[a + b*x]^3/3)/b

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Maple [A]
time = 0.15, size = 24, normalized size = 0.92

method	result	size
derivativedivides	\(-\frac {\left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (b x +a \right )\right )}{3}\right ) \tan \left (b x +a \right )}{b}\)	\(24\)
default	\(-\frac {\left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (b x +a \right )\right )}{3}\right ) \tan \left (b x +a \right )}{b}\)	\(24\)
risch	\(\frac {4 i \left (3 \,{\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{3}}\)	\(33\)
norman	\(\frac {-\frac {2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}+\frac {4 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {2 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 22, normalized size = 0.85 \begin {gather*} \frac {\tan \left (b x + a\right )^{3} + 3 \, \tan \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(tan(b*x + a)^3 + 3*tan(b*x + a))/b

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Fricas [A]
time = 2.47, size = 31, normalized size = 1.19 \begin {gather*} \frac {{\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{3 \, b \cos \left (b x + a\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*cos(b*x + a)^2 + 1)*sin(b*x + a)/(b*cos(b*x + a)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^{4}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4,x)

[Out]

Integral(sec(a + b*x)**4, x)

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Giac [A]
time = 0.44, size = 22, normalized size = 0.85 \begin {gather*} \frac {\tan \left (b x + a\right )^{3} + 3 \, \tan \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(tan(b*x + a)^3 + 3*tan(b*x + a))/b

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Mupad [B]
time = 0.08, size = 21, normalized size = 0.81 \begin {gather*} \frac {\mathrm {tan}\left (a+b\,x\right )\,\left ({\mathrm {tan}\left (a+b\,x\right )}^2+3\right )}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(a + b*x)^4,x)

[Out]

(tan(a + b*x)*(tan(a + b*x)^2 + 3))/(3*b)

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