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

Optimal. Leaf size=167 \[ -\frac {3 A \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+1);\frac {1}{6} (3 m+7);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+4);\frac {1}{6} (3 m+10);\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

[Out]

-3*A*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/6+1/2*m],[7/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(1+3*m)/(b*cos(d*x+c))
^(2/3)/(sin(d*x+c)^2)^(1/2)-3*B*cos(d*x+c)^(2+m)*hypergeom([1/2, 2/3+1/2*m],[5/3+1/2*m],cos(d*x+c)^2)*sin(d*x+
c)/d/(4+3*m)/(b*cos(d*x+c))^(2/3)/(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 = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {20, 2748, 2643} \[ -\frac {3 A \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+1);\frac {1}{6} (3 m+7);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+4);\frac {1}{6} (3 m+10);\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 140, normalized size = 0.84 \[ -\frac {3 \sqrt {\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+1}(c+d x) \left (A (3 m+4) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+1);\frac {1}{6} (3 m+7);\cos ^2(c+d x)\right )+B (3 m+1) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+4);\frac {m}{2}+\frac {5}{3};\cos ^2(c+d x)\right )\right )}{d (3 m+1) (3 m+4) (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +B \cos \left (d x +c \right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^m\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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