\(\int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx\) [709]

Optimal result

Integrand size = 23, antiderivative size = 307 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=-\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {\left (3 a^2-2 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{\sqrt {2} b^2 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {a \left (3 a^2-4 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{\sqrt {2} b^2 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \]

[Out]

-3/2*a^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(2/3)+1/2*(3*a^2-2*b^2)*AppellF1(1/2,-1/3,1/2,3/2,b*(1-sec(
d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^(1/3)*tan(d*x+c)/b^2/(a^2-b^2)/d/((a+b*sec(d*x+c))/(a+b))^(
1/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)-1/2*a*(3*a^2-4*b^2)*AppellF1(1/2,2/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*
sec(d*x+c))*((a+b*sec(d*x+c))/(a+b))^(2/3)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(2/3)*2^(1/2)/(1+sec(d*
x+c))^(1/2)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3924, 4092, 3919, 144, 143} \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=-\frac {a \left (3 a^2-4 b^2\right ) \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{\sqrt {2} b^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}+\frac {\left (3 a^2-2 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{\sqrt {2} b^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}} \]

[In]

Int[Sec[c + d*x]^3/(a + b*Sec[c + d*x])^(5/3),x]

[Out]

(-3*a^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(2/3)) + ((3*a^2 - 2*b^2)*AppellF1[1/2, 1/2, -1/
3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(Sqrt[2
]*b^2*(a^2 - b^2)*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) - (a*(3*a^2 - 4*b^2)*AppellF1
[1/2, 1/2, 2/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*x])/(a + b))^(2/3
)*Tan[c + d*x])/(Sqrt[2]*b^2*(a^2 - b^2)*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(2/3))

Rule 143

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

Rule 144

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

Rule 3919

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rule 3924

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)*Cot[e + f*
x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e +
f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[a*b*(m + 1) - (a^2 + b^2*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a,
b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4092

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

Rubi steps \begin{align*} \text {integral}& = -\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {3 \int \frac {\sec (c+d x) \left (-\frac {2 a b}{3}-\frac {1}{3} \left (3 a^2-2 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b \left (a^2-b^2\right )} \\ & = -\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {\left (a \left (3 a^2-4 b^2\right )\right ) \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac {\left (3 a^2-2 b^2\right ) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )} \\ & = -\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {\left (a \left (3 a^2-4 b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{2/3}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (\left (3 a^2-2 b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = -\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {\left (\left (3 a^2-2 b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}}}+\frac {\left (a \left (3 a^2-4 b^2\right ) \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \\ & = -\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {\left (3 a^2-2 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{\sqrt {2} b^2 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {a \left (3 a^2-4 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{\sqrt {2} b^2 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(19126\) vs. \(2(307)=614\).

Time = 44.10 (sec) , antiderivative size = 19126, normalized size of antiderivative = 62.30 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]

[In]

Integrate[Sec[c + d*x]^3/(a + b*Sec[c + d*x])^(5/3),x]

[Out]

Result too large to show

Maple [F]

\[\int \frac {\sec \left (d x +c \right )^{3}}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}}d x\]

[In]

int(sec(d*x+c)^3/(a+b*sec(d*x+c))^(5/3),x)

[Out]

int(sec(d*x+c)^3/(a+b*sec(d*x+c))^(5/3),x)

Fricas [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(a+b*sec(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{3}}}\, dx \]

[In]

integrate(sec(d*x+c)**3/(a+b*sec(d*x+c))**(5/3),x)

[Out]

Integral(sec(c + d*x)**3/(a + b*sec(c + d*x))**(5/3), x)

Maxima [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(a+b*sec(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/3), x)

Giac [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(a+b*sec(d*x+c))^(5/3),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]

[In]

int(1/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/3)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/3)), x)