\(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\) [18]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 (A+4 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 b d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4130, 3857, 2722} \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}-\frac {3 (A+4 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{4 b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[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 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 3857

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 4130

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]

Rubi steps \begin{align*} \text {integral}& = \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) \operatorname {Hypergeometric2F1}\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 [A] (verified)

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \cot (c+d x) \left (A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right )-2 C \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{2/3} \sqrt {-\tan ^2(c+d x)}}{4 b^2 d} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {A +C \sec \left (d x +c \right )^{2}}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]

[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)

Fricas [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

[In]

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

[Out]

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