3.18 \(\int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=90 \[ \frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}-\frac{3 (A+4 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{2},\frac{7}{6},\cos ^2(c+d x)\right )}{4 b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[Out]

(-3*(A + 4*C)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(4*b*d*(b*Sec[c + d*x])^(1/3)*Sqr
t[Sin[c + d*x]^2]) + (3*A*Tan[c + d*x])/(4*d*(b*Sec[c + d*x])^(4/3))

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Rubi [A]  time = 0.0707745, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4045, 3772, 2643} \[ \frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}-\frac{3 (A+4 C) \sin (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right )}{4 b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(4/3),x]

[Out]

(-3*(A + 4*C)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(4*b*d*(b*Sec[c + d*x])^(1/3)*Sqr
t[Sin[c + d*x]^2]) + (3*A*Tan[c + d*x])/(4*d*(b*Sec[c + d*x])^(4/3))

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx &=\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}+\frac{(A+4 C) \int (b \sec (c+d x))^{2/3} \, dx}{4 b^2}\\ &=\frac{3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}+\frac{\left ((A+4 C) \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{2/3}} \, dx}{4 b^2}\\ &=-\frac{3 (A+4 C) \cos (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\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 \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}\\ \end{align*}

Mathematica [C]  time = 0.433661, size = 124, normalized size = 1.38 \[ -\frac{3 i e^{-i (c+d x)} \left (2 (A+4 C) e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{2}{3},\frac{4}{3},-e^{2 i (c+d x)}\right )+A \left (-1+e^{4 i (c+d x)}\right )\right )}{8 b d \left (1+e^{2 i (c+d x)}\right ) \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(4/3),x]

[Out]

(((-3*I)/8)*(A*(-1 + E^((4*I)*(c + d*x))) + 2*(A + 4*C)*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(2/3)*Hy
pergeometric2F1[1/3, 2/3, 4/3, -E^((2*I)*(c + d*x))]))/(b*d*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))*(b*Sec[c
+ d*x])^(1/3))

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Maple [F]  time = 0.121, size = 0, normalized size = 0. \begin{align*} \int{(A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2}) \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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x)

[Out]

int((A+C*sec(d*x+c)^2)/(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{C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + 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 (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + 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 + C \sec ^{2}{\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+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(4/3),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(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{C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(4/3), x)