3.9.96 \(\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [896]

Optimal. Leaf size=116 \[ -\frac {3 A (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/4*A*(b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)-3/7*B*(
b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 2827, 2722} \begin {gather*} -\frac {3 A \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx &=b \int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\\ &=(A b) \int \sqrt [3]{b \cos (c+d x)} \, dx+B \int (b \cos (c+d x))^{4/3} \, dx\\ &=-\frac {3 A (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 87, normalized size = 0.75 \begin {gather*} -\frac {3 b \sqrt [3]{b \cos (c+d x)} \cot (c+d x) \left (7 A \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )+4 B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{28 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(b*Cos[c + d*x])^(1/3)*Cot[c + d*x]*(7*A*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2] + 4*B*Cos[c +
d*x]*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(28*d)

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )\right ) \sec \left (d x +c \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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