Integrand size = 23, antiderivative size = 378 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=-\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (9 a^2-7 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{2 \sqrt {2} b^3 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{2 \sqrt {2} b^3 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \]
-3/2*a^2*sec(d*x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(2/3)+3/4*(3 *a^2-b^2)*(a+b*sec(d*x+c))^(1/3)*tan(d*x+c)/b^2/(a^2-b^2)/d-1/4*a*(9*a^2-7 *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^3/(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)+1/4*(9*a^4-10*a^2*b^2-b^4)*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^3/(a^2-b^2)/d/(a+b*sec(d*x+c))^(2/3)*2^(1/2)/( 1+sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(21987\) vs. \(2(378)=756\).
Time = 44.23 (sec) , antiderivative size = 21987, normalized size of antiderivative = 58.17 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]
Time = 1.18 (sec) , antiderivative size = 367, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4332, 27, 3042, 4570, 27, 3042, 4495, 3042, 4321, 156, 155}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx\) |
| \(\Big \downarrow \) 4332 |
| \(\displaystyle -\frac {3 \int \frac {\sec (c+d x) \left (3 a^2-2 b \sec (c+d x) a-2 \left (3 a^2-b^2\right ) \sec ^2(c+d x)\right )}{3 (a+b \sec (c+d x))^{2/3}}dx}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec (c+d x) \left (3 a^2-2 b \sec (c+d x) a-2 \left (3 a^2-b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{2/3}}dx}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a^2-2 b \csc \left (c+d x+\frac {\pi }{2}\right ) a-2 \left (3 a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle -\frac {\frac {3 \int \frac {2 \sec (c+d x) \left (b \left (3 a^2+b^2\right )+a \left (9 a^2-7 b^2\right ) \sec (c+d x)\right )}{3 (a+b \sec (c+d x))^{2/3}}dx}{4 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (3 a^2+b^2\right )+a \left (9 a^2-7 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}}dx}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (3 a^2+b^2\right )+a \left (9 a^2-7 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 4495 |
| \(\displaystyle -\frac {\frac {\frac {a \left (9 a^2-7 b^2\right ) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)}dx}{b}-\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}}dx}{b}}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {a \left (9 a^2-7 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{b}}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 4321 |
| \(\displaystyle -\frac {\frac {\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}-\frac {a \left (9 a^2-7 b^2\right ) \tan (c+d x) \int \frac {\sqrt [3]{a+b \sec (c+d x)}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle -\frac {\frac {\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{2/3}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac {a \left (9 a^2-7 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} \int \frac {\sqrt [3]{\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle -\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}-\frac {\frac {\frac {\sqrt {2} a \left (9 a^2-7 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {\sqrt {2} \left (9 a^4-10 a^2 b^2-b^4\right ) \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}}{2 b}-\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{2 b d}}{2 b \left (a^2-b^2\right )}\) |
(-3*a^2*Sec[c + d*x]*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x]) ^(2/3)) - ((-3*(3*a^2 - b^2)*(a + b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(2*b *d) + ((Sqrt[2]*a*(9*a^2 - 7*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])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3 )) - (Sqrt[2]*(9*a^4 - 10*a^2*b^2 - b^4)*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])/(b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x] )^(2/3)))/(2*b))/(2*b*(a^2 - b^2))
3.8.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[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] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[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*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-a^2)*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^( m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[d^3/ (b*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x]) ^(n - 3)*Simp[a^2*(n - 3) + a*b*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*( m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n , 2]))
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b Int[ Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] + Simp[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] && N eQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \frac {\sec \left (d x +c \right )^{4}}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}}d x\]
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
integral((b*sec(d*x + c) + a)^(1/3)*sec(d*x + c)^4/(b^2*sec(d*x + c)^2 + 2 *a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{3}}}\, dx \]
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]