3.272 \(\int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=264 \[ -\frac {\left (a^2 (1-m)-b^2 (m+2)\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m-1}{2};\frac {m+1}{2};\cos ^2(c+d x)\right )}{d (1-m) m (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x)}{d m (m+2)}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\cos ^2(c+d x)\right )}{d m (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x)}{d \left (m^2+3 m+2\right )}-\frac {\sin (c+d x) \cos ^{m-1}(c+d x) (a \cos (c+d x)+b)^2}{d (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.76, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4397, 2889, 3050, 3033, 3023, 2748, 2643} \[ -\frac {\left (a^2 (1-m)-b^2 (m+2)\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m-1}{2};\frac {m+1}{2};\cos ^2(c+d x)\right )}{d (1-m) m (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x)}{d m (m+2)}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\cos ^2(c+d x)\right )}{d m (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x)}{d \left (m^2+3 m+2\right )}-\frac {\sin (c+d x) \cos ^{m-1}(c+d x) (a \cos (c+d x)+b)^2}{d (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

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

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Mathematica [C]  time = 30.60, size = 6669, normalized size = 25.26 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 1.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) \tan \left (d x + c\right ) - b^{2} \tan \left (d x + c\right )^{2} - a^{2}\right )} \cos \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-(a^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c)*tan(d*x + c) - b^2*tan(d*x + c)^2 - a^2)*cos(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + b*tan(d*x + c))^2*cos(d*x + c)^m, x)

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maple [F]  time = 2.25, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a \sin \left (d x +c \right )+b \tan \left (d x +c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^2,x)

[Out]

int(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(d*x + c) + b*tan(d*x + c))^2*cos(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^m*(a*sin(c + d*x) + b*tan(c + d*x))^2,x)

[Out]

int(cos(c + d*x)^m*(a*sin(c + d*x) + b*tan(c + d*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**m*(a*sin(d*x+c)+b*tan(d*x+c))**2,x)

[Out]

Integral((a*sin(c + d*x) + b*tan(c + d*x))**2*cos(c + d*x)**m, x)

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