\(\int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx\) [815]

Optimal result

Integrand size = 23, antiderivative size = 149 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=-\frac {6 b \left (5 a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-6/5*b*(5*a^2+b^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))/d+2*a
*(a^2+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+8/5*a*b^2*s
in(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*b^2*(a+b*sec(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(3/2)+6/5*b*(5*a^2+b^2)*sin(d*x+
c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4349, 3927, 4132, 3853, 3856, 2719, 4131, 2720} \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 a \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b \left (5 a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b^2 \sin (c+d x) (a+b \sec (c+d x))}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(-6*b*(5*a^2 + b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2])/d + (8*a*b^
2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)) + (6*b*(5*a^2 + b^2)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^2*
(a + b*Sec[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2))

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 3927

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Dist[1/(d*(m + n - 1)),
Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2)
+ 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Int
egerQ[m])

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 4132

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

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 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx \\ & = \frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \left (\frac {1}{2} a \left (5 a^2+b^2\right )+\frac {3}{2} b \left (5 a^2+b^2\right ) \sec (c+d x)+6 a b^2 \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \left (\frac {1}{2} a \left (5 a^2+b^2\right )+6 a b^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (3 b \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\left (a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (3 b \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\left (a \left (a^2+b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (3 b \left (5 a^2+b^2\right )\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {6 b \left (5 a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\frac {-6 b \left (5 a^2+b^2\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 a \left (a^2+b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+10 a b^2 \sin (c+d x)+3 \left (5 a^2 b+b^3\right ) \sin (2 (c+d x))+2 b^3 \tan (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(-6*b*(5*a^2 + b^2)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*a*(a^2 + b^2)*Cos[c + d*x]^(3/2)*Ellipti
cF[(c + d*x)/2, 2] + 10*a*b^2*Sin[c + d*x] + 3*(5*a^2*b + b^3)*Sin[2*(c + d*x)] + 2*b^3*Tan[c + d*x])/(5*d*Cos
[c + d*x]^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(187)=374\).

Time = 17.43 (sec) , antiderivative size = 711, normalized size of antiderivative = 4.77

method	result	size
default	\(\text {Expression too large to display}\)	\(711\)

[In]

int((a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),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^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))+
2/5*b^3/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(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*sin(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)+6*a*b^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)))+6*a^2*b/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(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)/(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 = 244, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{3} + i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} - i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, a b^{2} \cos \left (d x + c\right ) + b^{3} + 3 \, {\left (5 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(5*sqrt(2)*(I*a^3 + I*a*b^2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5
*sqrt(2)*(-I*a^3 - I*a*b^2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(
2)*(5*I*a^2*b + I*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c))) + 3*sqrt(2)*(-5*I*a^2*b - I*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(d*x + c) - I*sin(d*x + c))) - 2*(5*a*b^2*cos(d*x + c) + b^3 + 3*(5*a^2*b + b^3)*cos(d*x + c)^2)*sqrt(cos(d*x
+ c))*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

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

[In]

integrate((a+b*sec(d*x+c))**3/cos(d*x+c)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.90 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*b^3*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d
*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (6*a^2*b*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2
))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*a*b^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d
*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))