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(f*x+e))^m/f/(1-m)/m+(b^2-a^2*(1-m))*cos(f*x+e)*(d*cos(f*x+e))^m*hypergeom([1/2, 1/2+1/2*m],[
3/2+1/2*m],cos(f*x+e)^2)*sin(f*x+e)/f/(-m^2+1)/(sin(f*x+e)^2)^(1/2)+b*(d*cos(f*x+e))^m*(a+b*tan(f*x+e))/f/(1-m
)

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Rubi [A]  time = 0.24, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 3.75, 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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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \left (d \cos \left (f x + e\right )\right )^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \]

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)

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maple [F]  time = 1.26, size = 0, normalized size = 0.00 \[ \int \left (d \cos \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\cos \left (e+f\,x\right )\right )}^m\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cos {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]

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]

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

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