3.24 \(\int \frac{A+B \sec (c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=114 \[ -\frac{3 A b \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{6},\frac{13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac{3 B \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

[Out]

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

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Rubi [A]  time = 0.0890296, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3787, 3772, 2643} \[ -\frac{3 A b \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac{3 B \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[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]

Rubi steps

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

Mathematica [A]  time = 0.144719, size = 87, normalized size = 0.76 \[ -\frac{3 \sqrt{-\tan ^2(c+d x)} \csc (c+d x) \left (A \cos (c+d x) \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{1}{2},\frac{1}{3},\sec ^2(c+d x)\right )+4 B \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{5}{6},\sec ^2(c+d x)\right )\right )}{4 d (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{(A+B\sec \left ( dx+c \right ) ) \left ( b\sec \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{2}{3}}}{b^{2} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\left (c + d x \right )}}{\left (b \sec{\left (c + d x \right )}\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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