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

Optimal. Leaf size=396 \[ \frac{3^{3/4} (A-4 C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [3]{2} a d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}+\frac{3 \sqrt{2} A \tan (c+d x) F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{a d \sqrt{1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}-\frac{3 (A+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^{4/3}} \]

[Out]

(-3*(A + C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^(4/3)) + (3*Sqrt[2]*A*AppellF1[1/6, 1/2, 1, 7/6, (1 + Sec[
c + d*x])/2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(a*d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1/3)) + (3^(3/4
)*(A - 4*C)*EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3])*(1 +
Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(2^(1/3) - (1 + Sec[c + d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[
c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x]
)/(5*2^(1/3)*a*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(1/3)*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1
+ Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2)])

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Rubi [A]  time = 0.45772, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4053, 3924, 3779, 3778, 136, 3828, 3827, 63, 225} \[ \frac{3 \sqrt{2} A \tan (c+d x) F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{a d \sqrt{1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}-\frac{3 (A+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^{4/3}}+\frac{3^{3/4} (A-4 C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5 \sqrt [3]{2} a d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(A + C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^(4/3)) + (3*Sqrt[2]*A*AppellF1[1/6, 1/2, 1, 7/6, (1 + Sec[
c + d*x])/2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(a*d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1/3)) + (3^(3/4
)*(A - 4*C)*EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3])*(1 +
Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(2^(1/3) - (1 + Sec[c + d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[
c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x]
)/(5*2^(1/3)*a*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(1/3)*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1
+ Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2)])

Rule 4053

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

Rule 3924

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, I
nt[(a + b*Csc[e + f*x])^m, x], x] + Dist[d, Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[2*m]

Rule 3779

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 3778

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^n*Cot[c + d*x])/(d*Sqrt[1 + Csc[c + d*x]
]*Sqrt[1 - Csc[c + d*x]]), Subst[Int[(1 + (b*x)/a)^(n - 1/2)/(x*Sqrt[1 - (b*x)/a]), x], x, Csc[c + d*x]], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 3827

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx &=-\frac{3 (A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^{4/3}}-\frac{3 \int \frac{-\frac{5 a A}{3}+\frac{1}{3} a (A-4 C) \sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{5 a^2}\\ &=-\frac{3 (A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^{4/3}}+\frac{A \int \frac{1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{a}-\frac{(A-4 C) \int \frac{\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac{3 (A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^{4/3}}+\frac{\left (A \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac{1}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{a \sqrt [3]{a+a \sec (c+d x)}}-\frac{\left ((A-4 C) \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{5 a \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{3 (A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^{4/3}}-\frac{(A \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{a d \sqrt{1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}+\frac{((A-4 C) \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{5 a d \sqrt{1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{3 (A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^{4/3}}+\frac{3 \sqrt{2} A F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{a d \sqrt{1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}+\frac{(6 (A-4 C) \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{5 a d \sqrt{1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{3 (A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^{4/3}}+\frac{3 \sqrt{2} A F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{a d \sqrt{1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}+\frac{3^{3/4} (A-4 C) F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt{\frac{2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{5 \sqrt [3]{2} a d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt{-\frac{\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}\\ \end{align*}

Mathematica [F]  time = 3.13432, size = 0, normalized size = 0. \[ \int \frac{A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.161, size = 0, normalized size = 0. \begin{align*} \int{(A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2}) \left ( a+a\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)/(a+a*sec(d*x+c))^(4/3),x)

[Out]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

Timed out

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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 (a \left (\sec{\left (c + d x \right )} + 1\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)/(a+a*sec(d*x+c))**(4/3),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(a*(sec(c + d*x) + 1))**(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 (a \sec \left (d x + c\right ) + a\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)/(a+a*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

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