3.3.10 \(\int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [210]

Optimal. Leaf size=167 \[ -\frac {3 B \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {1}{6} (7+3 m);\frac {1}{6} (13+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \cos ^{3+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {1}{6} (10+3 m);\frac {1}{6} (16+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (10+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3*B*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/
(7+3*m)/(sin(d*x+c)^2)^(1/2)-3*C*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 = 0.09, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {20, 3089, 2827, 2722} \begin {gather*} -\frac {3 B \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) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(1/3)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-3*B*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*(7 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*C*Cos[c + d*x]^(3 + m)*(b*Cos[c + d*x])^(1/3)*Hype
rgeometric2F1[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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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 3089

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rubi steps

\begin {align*} \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {\sqrt [3]{b \cos (c+d x)} \int \cos ^{\frac {1}{3}+m}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt [3]{\cos (c+d x)}}\\ &=\frac {\sqrt [3]{b \cos (c+d x)} \int \cos ^{\frac {4}{3}+m}(c+d x) (B+C \cos (c+d x)) \, dx}{\sqrt [3]{\cos (c+d x)}}\\ &=\frac {\left (B \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \, dx}{\sqrt [3]{\cos (c+d x)}}+\frac {\left (C \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 {3 B \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 (7+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \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 = 0.41, size = 140, normalized size = 0.84 \begin {gather*} -\frac {3 \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \left (C (7+3 m) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2};\frac {8}{3}+\frac {m}{2};\cos ^2(c+d x)\right )+B (10+3 m) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (7+3 m);\frac {1}{6} (13+3 m);\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (7+3 m) (10+3 m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(1/3)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

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

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Maple [F]
time = 0.26, size = 0, normalized size = 0.00 \[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(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))*(b*cos(d*x + c))^(1/3)*cos(d*x + c)^m, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{b \cos {\left (c + d x \right )}} \left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*cos(c + d*x))**(1/3)*(B + C*cos(c + d*x))*cos(c + d*x)*cos(c + d*x)**m, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(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))*(b*cos(d*x + c))^(1/3)*cos(d*x + c)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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