3.10.32 \(\int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) [932]

Optimal. Leaf size=171 \[ \frac {3 A \cos ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-1+3 m);\frac {1}{6} (5+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{1+m}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2+3 m);\frac {1}{6} (8+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*A*cos(d*x+c)^m*hypergeom([1/2, -1/6+1/2*m],[5/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/b/d/(1-3*m)/(b*cos(d*x+c))^(
1/3)/(sin(d*x+c)^2)^(1/2)-3*B*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/3+1/2*m],[4/3+1/2*m],cos(d*x+c)^2)*sin(d*x+c)
/b/d/(2+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 171, 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, 2827, 2722} \begin {gather*} \frac {3 A \sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m-1);\frac {1}{6} (3 m+5);\cos ^2(c+d x)\right )}{b d (1-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+2);\frac {1}{6} (3 m+8);\cos ^2(c+d x)\right )}{b d (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx &=\frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{-\frac {4}{3}+m}(c+d x) (A+B \cos (c+d x)) \, dx}{b \sqrt [3]{b \cos (c+d x)}}\\ &=\frac {\left (A \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {4}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}}+\frac {\left (B \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}}\\ &=\frac {3 A \cos ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-1+3 m);\frac {1}{6} (5+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{1+m}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2+3 m);\frac {1}{6} (8+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 140, normalized size = 0.82 \begin {gather*} -\frac {3 \cos ^{1+m}(c+d x) \csc (c+d x) \left (A (2+3 m) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-1+3 m);\frac {1}{6} (5+3 m);\cos ^2(c+d x)\right )+B (-1+3 m) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2+3 m);\frac {1}{6} (8+3 m);\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-1+3 m) (2+3 m) (b \cos (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +B \cos \left (d x +c \right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

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))^(4/3),x)

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {4}{3}}}\, dx \end {gather*}

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))**(4/3),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )}^{4/3}} \,d x \end {gather*}

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))^(4/3),x)

[Out]

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

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