3.5.79 \(\int \sec ^2(e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [479]

Optimal. Leaf size=31 \[ \frac {\tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)} \]

[Out]

tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+1)

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Rubi [A]
time = 0.06, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3740, 2687, 32} \begin {gather*} \frac {\tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 3740

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Tan[e + f*x
])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps

\begin {align*} \int \sec ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx &=\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \sec ^2(e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\frac {\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \text {Subst}\left (\int (c x)^{n p} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 31, normalized size = 1.00 \begin {gather*} \frac {\tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 4.16, size = 10286, normalized size = 331.81

method	result	size
risch	\(\text {Expression too large to display}\)	\(10286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.32, size = 37, normalized size = 1.19 \begin {gather*} \frac {b^{p} c^{n p} {\left (\tan \left (f x + e\right )^{n}\right )}^{p} \tan \left (f x + e\right )}{{\left (n p + 1\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

b^p*c^(n*p)*(tan(f*x + e)^n)^p*tan(f*x + e)/((n*p + 1)*f)

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Fricas [A]
time = 2.27, size = 53, normalized size = 1.71 \begin {gather*} \frac {e^{\left (n p \log \left (\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + p \log \left (b\right )\right )} \sin \left (f x + e\right )}{{\left (f n p + f\right )} \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

e^(n*p*log(c*sin(f*x + e)/cos(f*x + e)) + p*log(b))*sin(f*x + e)/((f*n*p + f)*cos(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*sec(e + f*x)**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 1.11Unable to divide, perhaps due to rounding error%%%{1,[0,1,0,0]%%%} / %%%{1,[0,0,1,1]%%
%} Error: B

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Mupad [B]
time = 13.29, size = 31, normalized size = 1.00 \begin {gather*} \frac {\mathrm {tan}\left (e+f\,x\right )\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{f\,\left (n\,p+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^2,x)

[Out]

(tan(e + f*x)*(b*(c*tan(e + f*x))^n)^p)/(f*(n*p + 1))

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