\(\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx\) [173]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {16 a^2 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {16 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

4/3*a^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d+16/5*a^2*sin(d*x+c)*sec(d*x+c)^(1/
2)/d-16/5*a^2*(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))*cos(d*x+c)
^(1/2)*sec(d*x+c)^(1/2)/d+4/3*a^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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 161, 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 = {3873, 3853, 3856, 2720, 4131, 2719} \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {2 a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {4 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {16 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {16 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

[In]

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

[Out]

(-16*a^2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (16*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (4*a^2*Se
c[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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]

Rubi steps \begin{align*} \text {integral}& = \left (2 a^2\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx+\int \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx \\ & = \frac {4 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} \left (2 a^2\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (8 a^2\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {16 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \left (8 a^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {4 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {16 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \left (8 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {16 a^2 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {16 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.06 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.67 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^2 \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^2(c+d x) \left (12 \left (1+e^{2 i (c+d x)}\right )+12 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+5 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {24 \cos (d x) \csc (c)+(10+3 \sec (c+d x)) \tan (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}\right )}{30 d} \]

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*
Cos[c + d*x]^2*(12*(1 + E^((2*I)*(c + d*x))) + 12*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeomet
ric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x
))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (24*Cos[d*
x]*Csc[c] + (10 + 3*Sec[c + d*x])*Tan[c + d*x])/Sec[c + d*x]^(3/2)))/(30*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(189)=378\).

Time = 26.28 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.40

method	result	size
default	\(-\frac {a^{2} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {68 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {32 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {16 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{10 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(386\)
parts	\(\text {Expression too large to display}\)	\(761\)

[In]

int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/3*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+68/15*(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))-
32/5*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-16/5*(s
in(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-1/10*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)^3)/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 = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.25 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (24 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-2/15*(5*I*sqrt(2)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*
a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*I*sqrt(2)*a^2*cos(d*x + c)^2
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*I*sqrt(2)*a^2*cos(d*x
+ c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (24*a^2*cos(d*x + c
)^2 + 10*a^2*cos(d*x + c) + 3*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

[In]

int((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(3/2),x)

[Out]

int((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(3/2), x)