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

Optimal. Leaf size=155 \[ \frac{\left (b^2-a^2 (1-m)\right ) \sin (e+f x) \cos (e+f x) (d \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt{\sin ^2(e+f x)}}-\frac{a b (2-m) (d \cos (e+f x))^m}{f (1-m) m}+\frac{b (a+b \tan (e+f x)) (d \cos (e+f x))^m}{f (1-m)} \]

[Out]

-((a*b*(2 - m)*(d*Cos[e + f*x])^m)/(f*(1 - m)*m)) + ((b^2 - a^2*(1 - m))*Cos[e + f*x]*(d*Cos[e + f*x])^m*Hyper
geometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(1 - m)*(1 + m)*Sqrt[Sin[e + f*x]^2])
+ (b*(d*Cos[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(1 - m))

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Rubi [A]  time = 0.238488, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3515, 3508, 3486, 3772, 2643} \[ \frac{\left (b^2-a^2 (1-m)\right ) \sin (e+f x) \cos (e+f x) (d \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt{\sin ^2(e+f x)}}-\frac{a b (2-m) (d \cos (e+f x))^m}{f (1-m) m}+\frac{b (a+b \tan (e+f x)) (d \cos (e+f x))^m}{f (1-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*(2 - m)*(d*Cos[e + f*x])^m)/(f*(1 - m)*m)) + ((b^2 - a^2*(1 - m))*Cos[e + f*x]*(d*Cos[e + f*x])^m*Hyper
geometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(1 - m)*(1 + m)*Sqrt[Sin[e + f*x]^2])
+ (b*(d*Cos[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(1 - m))

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[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 \cos (e+f x))^m (a+b \tan (e+f x))^2 \, dx &=\left ((d \cos (e+f x))^m (d \sec (e+f x))^m\right ) \int (d \sec (e+f x))^{-m} (a+b \tan (e+f x))^2 \, dx\\ &=\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))}{f (1-m)}+\frac{\left ((d \cos (e+f x))^m (d \sec (e+f x))^m\right ) \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 \cos (e+f x))^m}{f (1-m) m}+\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))}{f (1-m)}+\frac{\left (\left (-b^2+a^2 (1-m)\right ) (d \cos (e+f x))^m (d \sec (e+f x))^m\right ) \int (d \sec (e+f x))^{-m} \, dx}{1-m}\\ &=-\frac{a b (2-m) (d \cos (e+f x))^m}{f (1-m) m}+\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))}{f (1-m)}+\frac{\left (\left (-b^2+a^2 (1-m)\right ) \left (\frac{\cos (e+f x)}{d}\right )^{-m} (d \cos (e+f x))^m\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^m \, dx}{1-m}\\ &=-\frac{a b (2-m) (d \cos (e+f x))^m}{f (1-m) m}+\frac{\left (b^2-a^2 (1-m)\right ) \cos (e+f x) (d \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{f (1-m) (1+m) \sqrt{\sin ^2(e+f x)}}+\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))}{f (1-m)}\\ \end{align*}

Mathematica [C]  time = 3.73468, size = 330, normalized size = 2.13 \[ \frac{\cos (e+f x) (a+b \tan (e+f x))^2 (d \cos (e+f x))^m \left (\sqrt{\sin ^2(e+f x)} \left (-\frac{a^2 \cos (e+f x) \cot (e+f x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{m+1}-\frac{b^2 \csc (e+f x) \, _2F_1\left (-\frac{1}{2},\frac{m-1}{2};\frac{m+1}{2};\cos ^2(e+f x)\right )}{m-1}\right )-\frac{a b 2^{1-m} \left (e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right )\right )^m \, _2F_1\left (1,\frac{m}{2};1-\frac{m}{2};-e^{2 i (e+f x)}\right ) \cos ^{1-m}(e+f x)}{m}+\frac{a b 2^{1-m} e^{2 i (e+f x)} \left (e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right )\right )^m \, _2F_1\left (1,\frac{m+2}{2};2-\frac{m}{2};-e^{2 i (e+f x)}\right ) \cos ^{1-m}(e+f x)}{m-2}\right )}{f (a \cos (e+f x)+b \sin (e+f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.371, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \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*cos(f*x+e))^m*(a+b*tan(f*x+e))^2,x)

[Out]

int((d*cos(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 \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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 \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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