3.441 \(\int (a+b \sec ^2(e+f x))^p (d \tan (e+f x))^m \, dx\)

Optimal. Leaf size=105 \[ \frac{(d \tan (e+f x))^{m+1} \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (\frac{m+1}{2};1,-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{d f (m+1)} \]

[Out]

(AppellF1[(1 + m)/2, 1, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*(d*Tan[e + f*x])^(1 + m
)*(a + b + b*Tan[e + f*x]^2)^p)/(d*f*(1 + m)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)

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Rubi [A]  time = 0.202328, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4141, 1975, 511, 510} \[ \frac{(d \tan (e+f x))^{m+1} \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (\frac{m+1}{2};1,-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{d f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^p*(d*Tan[e + f*x])^m,x]

[Out]

(AppellF1[(1 + m)/2, 1, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*(d*Tan[e + f*x])^(1 + m
)*(a + b + b*Tan[e + f*x]^2)^p)/(d*f*(1 + m)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(d x)^m \left (a+b \left (1+x^2\right )\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(d x)^m \left (a+b+b x^2\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{(d x)^m \left (1+\frac{b x^2}{a+b}\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{F_1\left (\frac{1+m}{2};1,-p;\frac{3+m}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right ) (d \tan (e+f x))^{1+m} \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}}{d f (1+m)}\\ \end{align*}

Mathematica [B]  time = 3.80146, size = 259, normalized size = 2.47 \[ \frac{\sin (e+f x) \cos (e+f x) (d \tan (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )}{f (m+1) \left (\frac{2 \tan ^2(e+f x) \left (b p F_1\left (\frac{m+3}{2};1-p,1;\frac{m+5}{2};-\frac{b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )-(a+b) F_1\left (\frac{m+3}{2};-p,2;\frac{m+5}{2};-\frac{b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )\right )}{(m+3) (a+b)}+F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^p*(d*Tan[e + f*x])^m,x]

[Out]

(AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f*x]*(a + b*Sec
[e + f*x]^2)^p*Sin[e + f*x]*(d*Tan[e + f*x])^m)/(f*(1 + m)*(AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*Tan[e +
f*x]^2)/(a + b)), -Tan[e + f*x]^2] + (2*(b*p*AppellF1[(3 + m)/2, 1 - p, 1, (5 + m)/2, -((b*Tan[e + f*x]^2)/(a
+ b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[(3 + m)/2, -p, 2, (5 + m)/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e
+ f*x]^2])*Tan[e + f*x]^2)/((a + b)*(3 + m))))

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Maple [F]  time = 0.804, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p} \left ( d\tan \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)^p*(d*tan(f*x+e))^m,x)

[Out]

int((a+b*sec(f*x+e)^2)^p*(d*tan(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*(d*tan(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*(d*tan(f*x + e))^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*(d*tan(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e)^2 + a)^p*(d*tan(f*x + e))^m, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)**2)**p*(d*tan(f*x+e))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*(d*tan(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*(d*tan(f*x + e))^m, x)