\(\int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\) [544]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (warning: unable to verify)
   Maple [B] (verified)
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F(-1)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 463 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=-\frac {2 a (a-b) \sqrt {a+b} \left (8 a^4+51 a^2 b^2+741 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{693 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4+6 a^3 b+57 a^2 b^2-606 a b^3+135 b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{693 b^3 d}+\frac {2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac {2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac {2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \]

[Out]

-2/693*a*(a-b)*(8*a^4+51*a^2*b^2+741*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b)
)^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d-2/693*(a-b)*(8*a^4+6
*a^3*b+57*a^2*b^2-606*a*b^3+135*b^4)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/
2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d+2/693*a*(8*a^2+67*b^2)*(a
+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^2/d+2/693*(8*a^2+81*b^2)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d-8/99*a*(a+b
*sec(d*x+c))^(7/2)*tan(d*x+c)/b^2/d+2/11*sec(d*x+c)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d+2/693*(8*a^4+57*a^2*
b^2+135*b^4)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3950, 4167, 4087, 4090, 3917, 4089} \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\frac {2 \left (8 a^2+81 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{693 b^2 d}+\frac {2 a \left (8 a^2+67 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{693 b^2 d}-\frac {2 a (a-b) \sqrt {a+b} \left (8 a^4+51 a^2 b^2+741 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{693 b^4 d}+\frac {2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{693 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4+6 a^3 b+57 a^2 b^2-606 a b^3+135 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{693 b^3 d}-\frac {8 a \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{99 b^2 d}+\frac {2 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d} \]

[In]

Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(-2*a*(a - b)*Sqrt[a + b]*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])
/(693*b^4*d) - (2*(a - b)*Sqrt[a + b]*(8*a^4 + 6*a^3*b + 57*a^2*b^2 - 606*a*b^3 + 135*b^4)*Cot[c + d*x]*Ellipt
icF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-
((b*(1 + Sec[c + d*x]))/(a - b))])/(693*b^3*d) + (2*(8*a^4 + 57*a^2*b^2 + 135*b^4)*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(693*b^2*d) + (2*a*(8*a^2 + 67*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(693*b^2*d) + (2*(8*a
^2 + 81*b^2)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*b^2*d) - (8*a*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*
x])/(99*b^2*d) + (2*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d)

Rule 3917

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

Rule 3950

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d^3)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + n - 1))), x] + Dist[d^3/(b*(m +
 n - 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 3)*Simp[a*(n - 3) + b*(m + n - 2)*Csc[e + f*x] - a*
(n - 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3] && (Integ
erQ[n] || IntegersQ[2*m, 2*n]) &&  !IGtQ[m, 2]

Rule 4087

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

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

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

Rule 4167

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_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
 + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {2 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (a+\frac {9}{2} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{11 b} \\ & = -\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (-\frac {5 a b}{2}+\frac {1}{4} \left (8 a^2+81 b^2\right ) \sec (c+d x)\right ) \, dx}{99 b^2} \\ & = \frac {2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {8 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {15}{8} b \left (2 a^2-27 b^2\right )+\frac {5}{8} a \left (8 a^2+67 b^2\right ) \sec (c+d x)\right ) \, dx}{693 b^2} \\ & = \frac {2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac {2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {16 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (-\frac {15}{8} a b \left (a^2-101 b^2\right )+\frac {15}{16} \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sec (c+d x)\right ) \, dx}{3465 b^2} \\ & = \frac {2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac {2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac {2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac {32 \int \frac {\sec (c+d x) \left (\frac {15}{32} b \left (2 a^4+663 a^2 b^2+135 b^4\right )+\frac {15}{32} a \left (8 a^4+51 a^2 b^2+741 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{10395 b^2} \\ & = \frac {2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac {2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac {2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}-\frac {\left ((a-b) \left (8 a^4+6 a^3 b+57 a^2 b^2-606 a b^3+135 b^4\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{693 b^2}+\frac {\left (a \left (8 a^4+51 a^2 b^2+741 b^4\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{693 b^2} \\ & = -\frac {2 a (a-b) \sqrt {a+b} \left (8 a^4+51 a^2 b^2+741 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{693 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4+6 a^3 b+57 a^2 b^2-606 a b^3+135 b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{693 b^3 d}+\frac {2 \left (8 a^4+57 a^2 b^2+135 b^4\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{693 b^2 d}+\frac {2 a \left (8 a^2+67 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{693 b^2 d}+\frac {2 \left (8 a^2+81 b^2\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 14.13 (sec) , antiderivative size = 615, normalized size of antiderivative = 1.33 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 a \left (8 a^5+8 a^4 b+51 a^3 b^2+51 a^2 b^3+741 a b^4+741 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (8 a^5+2 a^4 b+51 a^3 b^2+663 a^2 b^3+741 a b^4+135 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \left (8 a^4+51 a^2 b^2+741 b^4\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{693 b^3 d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2 a \left (8 a^4+51 a^2 b^2+741 b^4\right ) \sin (c+d x)}{693 b^3}+\frac {2}{693} \sec ^3(c+d x) \left (113 a^2 \sin (c+d x)+81 b^2 \sin (c+d x)\right )+\frac {2 \sec ^2(c+d x) \left (3 a^3 \sin (c+d x)+229 a b^2 \sin (c+d x)\right )}{693 b}+\frac {2 \sec (c+d x) \left (-4 a^4 \sin (c+d x)+205 a^2 b^2 \sin (c+d x)+135 b^4 \sin (c+d x)\right )}{693 b^2}+\frac {46}{99} a b \sec ^3(c+d x) \tan (c+d x)+\frac {2}{11} b^2 \sec ^4(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*a*(8*a^5 + 8*a^4*b + 51*a^3*b^2 + 51*a
^2*b^3 + 741*a*b^4 + 741*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co
s[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(8*a^5 + 2*a^4*b + 51*a^3*b^2 + 663*a
^2*b^3 + 741*a*b^4 + 135*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co
s[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(8*a^4 + 51*a^2*b^2 + 741*b^4)*Cos[c +
d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(693*b^3*d*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c
+ d*x)/2]^2]*Sec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*a*(8*a^4 + 51*a^2*b^2 + 741*
b^4)*Sin[c + d*x])/(693*b^3) + (2*Sec[c + d*x]^3*(113*a^2*Sin[c + d*x] + 81*b^2*Sin[c + d*x]))/693 + (2*Sec[c
+ d*x]^2*(3*a^3*Sin[c + d*x] + 229*a*b^2*Sin[c + d*x]))/(693*b) + (2*Sec[c + d*x]*(-4*a^4*Sin[c + d*x] + 205*a
^2*b^2*Sin[c + d*x] + 135*b^4*Sin[c + d*x]))/(693*b^2) + (46*a*b*Sec[c + d*x]^3*Tan[c + d*x])/99 + (2*b^2*Sec[
c + d*x]^4*Tan[c + d*x])/11))/(d*(b + a*Cos[c + d*x])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3642\) vs. \(2(421)=842\).

Time = 166.32 (sec) , antiderivative size = 3643, normalized size of antiderivative = 7.87

method result size
default \(\text {Expression too large to display}\) \(3643\)

[In]

int(sec(d*x+c)^4*(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/693/d/b^3*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(224*a*b^5*tan(d*x+c)*sec(d*x+c)^3-4*a^5*b*
cos(d*x+c)*sin(d*x+c)+51*a^4*b^2*cos(d*x+c)*sin(d*x+c)+205*a^3*b^3*cos(d*x+c)*sin(d*x+c)+741*a^2*b^4*cos(d*x+c
)*sin(d*x+c)+8*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^6+135*a*b^5*cos(d*x+c)*sin(d*x+c)+116*a^3*b^3*tan(d*x+c)*sec(d*x+c)+
274*a^2*b^4*tan(d*x+c)*sec(d*x+c)+310*a*b^5*tan(d*x+c)*sec(d*x+c)+274*a^2*b^4*tan(d*x+c)*sec(d*x+c)^2+224*a*b^
5*tan(d*x+c)*sec(d*x+c)^2-135*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^6+8*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*c
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^5*b*cos(d*x+c)^2+51*(c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c
),((a-b)/(a+b))^(1/2))*a^4*b^2*cos(d*x+c)^2+51*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b^3*cos(d*x+c)^2+741*(cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a
+b))^(1/2))*a^2*b^4*cos(d*x+c)^2+741*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^5*cos(d*x+c)^2-16*EllipticF(cot(d*x+c)-csc(d
*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a
^5*b*cos(d*x+c)-4*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a
+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^4*b^2*cos(d*x+c)-102*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b)
)^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b^3*cos(d*x+c)-
51*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*
x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b^3*cos(d*x+c)^2-663*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^4*cos(d*x+c)^2-741*Ellipti
cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos
(d*x+c)+1))^(1/2)*a*b^5*cos(d*x+c)^2+116*a^3*b^3*tan(d*x+c)+434*a^2*b^4*tan(d*x+c)+310*a*b^5*tan(d*x+c)+81*b^6
*tan(d*x+c)*sec(d*x+c)+81*b^6*tan(d*x+c)*sec(d*x+c)^2+63*b^6*tan(d*x+c)*sec(d*x+c)^3+8*a^6*cos(d*x+c)*sin(d*x+
c)+63*b^6*tan(d*x+c)*sec(d*x+c)^4+4*a^5*b*sin(d*x+c)-a^4*b^2*sin(d*x+c)+256*a^3*b^3*sin(d*x+c)+434*a^2*b^4*sin
(d*x+c)+876*a*b^5*sin(d*x+c)-a^4*b^2*tan(d*x+c)-1326*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^4*cos(d*x+c)-1482*Elliptic
F(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(
d*x+c)+1))^(1/2)*a*b^5*cos(d*x+c)+16*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^5*b*cos(d*x+c)+102*(cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^
4*b^2*cos(d*x+c)+102*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipti
cE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b^3*cos(d*x+c)+1482*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+
b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^4*cos(d*x
+c)+1482*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c
)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^5*cos(d*x+c)-8*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^5*b*cos(d*x+c)^2-2*EllipticF(c
ot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*a^4*b^2*cos(d*x+c)^2-135*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^6*cos(d*x+c)^2-8*EllipticF(cot(d*x+c)-csc(d*
x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^
5*b-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos
(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^4*b^2-51*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b^3-663*EllipticF(cot(d*x+c)-csc(d*x+c)
,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^
4-741*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos
(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^5+8*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)
+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^6*cos(d*x+c)^2-270*EllipticF(cot(d*x+c)-csc(
d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*
b^6*cos(d*x+c)+16*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(
cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^6*cos(d*x+c)+8*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2
))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^5*b+51*EllipticE(cot(d*
x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*a^4*b^2+51*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b^3+741*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^4+741*EllipticE(cot(d*
x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*a*b^5+135*b^6*sin(d*x+c)+135*b^6*tan(d*x+c))

Fricas [F]

\[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*sec(d*x + c)^6 + 2*a*b*sec(d*x + c)^5 + a^2*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a), x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

[In]

int((a + b/cos(c + d*x))^(5/2)/cos(c + d*x)^4,x)

[Out]

int((a + b/cos(c + d*x))^(5/2)/cos(c + d*x)^4, x)