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

Optimal. Leaf size=147 \[ \frac{d \left (b^2-a^2 (m+1)\right ) \sin (e+f x) (d \sec (e+f x))^{m-1} \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt{\sin ^2(e+f x)}}+\frac{a b (m+2) (d \sec (e+f x))^m}{f m (m+1)}+\frac{b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)} \]

[Out]

(a*b*(2 + m)*(d*Sec[e + f*x])^m)/(f*m*(1 + m)) + (d*(b^2 - a^2*(1 + m))*Hypergeometric2F1[1/2, (1 - m)/2, (3 -
m)/2, Cos[e + f*x]^2]*(d*Sec[e + f*x])^(-1 + m)*Sin[e + f*x])/(f*(1 - m)*(1 + m)*Sqrt[Sin[e + f*x]^2]) + (b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(1 + m))

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Rubi [A]  time = 0.166241, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3508, 3486, 3772, 2643} \[ \frac{d \left (b^2-a^2 (m+1)\right ) \sin (e+f x) (d \sec (e+f x))^{m-1} \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt{\sin ^2(e+f x)}}+\frac{a b (m+2) (d \sec (e+f x))^m}{f m (m+1)}+\frac{b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(2 + m)*(d*Sec[e + f*x])^m)/(f*m*(1 + m)) + (d*(b^2 - a^2*(1 + m))*Hypergeometric2F1[1/2, (1 - m)/2, (3 -
m)/2, Cos[e + f*x]^2]*(d*Sec[e + f*x])^(-1 + m)*Sin[e + f*x])/(f*(1 - m)*(1 + m)*Sqrt[Sin[e + f*x]^2]) + (b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(1 + m))

Rule 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx &=\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\frac{\int (d \sec (e+f x))^m \left (-b^2+a^2 (1+m)+a b (2+m) \tan (e+f x)\right ) \, dx}{1+m}\\ &=\frac{a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\left (a^2-\frac{b^2}{1+m}\right ) \int (d \sec (e+f x))^m \, dx\\ &=\frac{a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\left (\left (a^2-\frac{b^2}{1+m}\right ) \left (\frac{\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{-m} \, dx\\ &=\frac{a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}-\frac{\left (a^2-\frac{b^2}{1+m}\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};\cos ^2(e+f x)\right ) (d \sec (e+f x))^m \sin (e+f x)}{f (1-m) \sqrt{\sin ^2(e+f x)}}+\frac{b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}\\ \end{align*}

Mathematica [C]  time = 26.5429, size = 11095, normalized size = 75.48 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.325, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{m} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^2*(d*sec(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^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \left (d \sec \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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