3.363 \(\int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=232 \[ -\frac {3 b (A (3 m+10)+C (3 m+7)) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+7);\frac {1}{6} (3 m+13);\cos ^2(c+d x)\right )}{d (3 m+7) (3 m+10) \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+3}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+10);\frac {1}{6} (3 m+16);\cos ^2(c+d x)\right )}{d (3 m+10) \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x)}{d (3 m+10)} \]

[Out]

3*b*C*cos(d*x+c)^(2+m)*(b*cos(d*x+c))^(1/3)*sin(d*x+c)/d/(10+3*m)-3*b*(C*(7+3*m)+A*(10+3*m))*cos(d*x+c)^(2+m)*
(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 7/6+1/2*m],[13/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(9*m^2+51*m+70)/(sin(d
*x+c)^2)^(1/2)-3*b*B*cos(d*x+c)^(3+m)*(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 5/3+1/2*m],[8/3+1/2*m],cos(d*x+c)^2
)*sin(d*x+c)/d/(10+3*m)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 222, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {20, 3023, 2748, 2643} \[ -\frac {3 b \left (\frac {A}{3 m+7}+\frac {C}{3 m+10}\right ) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+7);\frac {1}{6} (3 m+13);\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+3}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+10);\frac {1}{6} (3 m+16);\cos ^2(c+d x)\right )}{d (3 m+10) \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x)}{d (3 m+10)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(3*b*C*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(d*(10 + 3*m)) - (3*b*(A/(7 + 3*m) + C/(10 +
3*m))*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*
x]^2]*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2]) - (3*b*B*Cos[c + d*x]^(3 + m)*(b*Cos[c + d*x])^(1/3)*Hypergeometr
ic2F1[1/2, (10 + 3*m)/6, (16 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(10 + 3*m)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + 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]

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 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]

Rubi steps

\begin {align*} \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {\left (b \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt [3]{\cos (c+d x)}}\\ &=\frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}+\frac {\left (3 b \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \left (\frac {1}{3} \left (3 C \left (\frac {7}{3}+m\right )+3 A \left (\frac {10}{3}+m\right )\right )+\frac {1}{3} B (10+3 m) \cos (c+d x)\right ) \, dx}{(10+3 m) \sqrt [3]{\cos (c+d x)}}\\ &=\frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}+\frac {\left (b B \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {7}{3}+m}(c+d x) \, dx}{\sqrt [3]{\cos (c+d x)}}+\frac {\left (b (C (7+3 m)+A (10+3 m)) \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \, dx}{(10+3 m) \sqrt [3]{\cos (c+d x)}}\\ &=\frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}-\frac {3 b \left (\frac {A}{7+3 m}+\frac {C}{10+3 m}\right ) \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {1}{6} (7+3 m);\frac {1}{6} (13+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \cos ^{3+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {1}{6} (10+3 m);\frac {1}{6} (16+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (10+3 m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 169, normalized size = 0.73 \[ -\frac {3 \sin (c+d x) (b \cos (c+d x))^{4/3} \cos ^{m+1}(c+d x) \left ((A (3 m+10)+C (3 m+7)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+7);\frac {1}{6} (3 m+13);\cos ^2(c+d x)\right )+B (3 m+7) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m}{2}+\frac {5}{3};\frac {m}{2}+\frac {8}{3};\cos ^2(c+d x)\right )-C (3 m+7) \sqrt {\sin ^2(c+d x)}\right )}{d (3 m+7) (3 m+10) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(4/3)*Sin[c + d*x]*(B*(7 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2,
5/3 + m/2, 8/3 + m/2, Cos[c + d*x]^2] + (C*(7 + 3*m) + A*(10 + 3*m))*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 +
3*m)/6, Cos[c + d*x]^2] - C*(7 + 3*m)*Sqrt[Sin[c + d*x]^2]))/(d*(7 + 3*m)*(10 + 3*m)*Sqrt[Sin[c + d*x]^2])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + B b \cos \left (d x + c\right )^{2} + A b \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b*cos(d*x + c)^3 + B*b*cos(d*x + c)^2 + A*b*cos(d*x + c))*(b*cos(d*x + c))^(1/3)*cos(d*x + c)^m, x
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3)*cos(d*x + c)^m, x)

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maple [F]  time = 0.55, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

int(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3)*cos(d*x + c)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^m*(b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^m*(b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**m*(b*cos(d*x+c))**(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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