\(\int \frac {(a+b \sec (c+d x))^2}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [809]

Optimal result

Integrand size = 23, antiderivative size = 160 \[ \int \frac {(a+b \sec (c+d x))^2}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {12 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 a^2+5 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 a b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {12 a b \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-12/5*a*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(7*a^2+
5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/7*b^2*sin(d*x
+c)/d/cos(d*x+c)^(7/2)+4/5*a*b*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*(7*a^2+5*b^2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+
12/5*a*b*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4349, 3873, 3853, 3856, 2719, 4131, 2720} \[ \int \frac {(a+b \sec (c+d x))^2}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (7 a^2+5 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 \left (7 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {12 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {12 a b \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

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

[Out]

(-12*a*b*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(7*a^2 + 5*b^2)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*b^2*Sin[
c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (4*a*b*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(7*a^2 + 5*b^2)*Sin[c
+ d*x])/(21*d*Cos[c + d*x]^(3/2)) + (12*a*b*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3853

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4131

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

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx \\ & = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right ) \, dx+\left (2 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {7}{2}}(c+d x) \, dx \\ & = \frac {2 b^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 a b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{7} \left (\left (7 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx \\ & = \frac {2 b^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 a b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {12 a b \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} \left (6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (\left (7 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {2 b^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 a b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {12 a b \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} (6 a b) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (7 a^2+5 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {12 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 a^2+5 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 a b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {12 a b \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b \sec (c+d x))^2}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-252 a b \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (7 a^2+5 b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+84 a b \sin (c+d x)+252 a b \cos ^2(c+d x) \sin (c+d x)+35 a^2 \sin (2 (c+d x))+25 b^2 \sin (2 (c+d x))+30 b^2 \tan (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

(-252*a*b*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10*(7*a^2 + 5*b^2)*Cos[c + d*x]^(5/2)*EllipticF[(c +
d*x)/2, 2] + 84*a*b*Sin[c + d*x] + 252*a*b*Cos[c + d*x]^2*Sin[c + d*x] + 35*a^2*Sin[2*(c + d*x)] + 25*b^2*Sin[
2*(c + d*x)] + 30*b^2*Tan[c + d*x])/(105*d*Cos[c + d*x]^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(688\) vs. \(2(192)=384\).

Time = 22.42 (sec) , antiderivative size = 689, normalized size of antiderivative = 4.31

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 a^{2} \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )+2 b^{2} \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{56 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{4}}-\frac {5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{42 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{21 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )+\frac {4 a b \left (24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(689\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
)+2*b^2*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1
/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^
2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2
*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+4/5*a*b/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*si
n(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(
1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b \sec (c+d x))^2}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-126 i \, \sqrt {2} a b \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 126 i \, \sqrt {2} a b \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 5 \, \sqrt {2} {\left (7 i \, a^{2} + 5 i \, b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-7 i \, a^{2} - 5 i \, b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (126 \, a b \cos \left (d x + c\right )^{3} + 42 \, a b \cos \left (d x + c\right ) + 5 \, {\left (7 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/105*(-126*I*sqrt(2)*a*b*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si
n(d*x + c))) + 126*I*sqrt(2)*a*b*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c)
- I*sin(d*x + c))) - 5*sqrt(2)*(7*I*a^2 + 5*I*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c)) - 5*sqrt(2)*(-7*I*a^2 - 5*I*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin
(d*x + c)) + 2*(126*a*b*cos(d*x + c)^3 + 42*a*b*cos(d*x + c) + 5*(7*a^2 + 5*b^2)*cos(d*x + c)^2 + 15*b^2)*sqrt
(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.71 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.71 \[ \int \frac {(a+b \sec (c+d x))^2}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {30\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(30*b^2*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2) + 70*a^2*cos(c + d*x)^2*sin(c + d*x)*hyperge
om([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 84*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d
*x)^2))/(105*d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))