\(\int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [515]

Optimal result

Integrand size = 23, antiderivative size = 291 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 \left (a^2-b^2\right ) d}-\frac {2 a \left (a^2-2 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 a^2-5 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{6 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-5 b^2\right )-4 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^2 d} \]

[Out]

-1/3*sec(d*x+c)^3*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)/d-1/6*sec(d*x+c)*(b*(a^2-5*b^2)-4*a*(a^2-2
*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^2/d+2/3*a*(a^2-2*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/s
in(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/(
a^2-b^2)^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/6*(4*a^2-5*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/
4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/(a^2
-b^2)/d/(a+b*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2775, 2945, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-5 b^2\right )-4 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^2}+\frac {\left (4 a^2-5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

[In]

Int[Sec[c + d*x]^4/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-1/3*(Sec[c + d*x]^3*(b - a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)*d) - (2*a*(a^2 - 2*b^2)*Ellip
ticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3*(a^2 - b^2)^2*d*Sqrt[(a + b*Sin[c + d*x])
/(a + b)]) + ((4*a^2 - 5*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])
/(6*(a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(b*(a^2 - 5*b^2) - 4*a*(a
^2 - 2*b^2)*Sin[c + d*x]))/(6*(a^2 - b^2)^2*d)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2831

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2945

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c -
b*d)*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

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

Mathematica [A] (verified)

Time = 2.90 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {16 a \left (a^3+a^2 b-2 a b^2-2 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-4 \left (4 a^4-9 a^2 b^2+5 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+\sec ^3(c+d x) \left (-4 a^3 b+10 a b^3+\left (-6 a^3 b+14 a b^3\right ) \cos (2 (c+d x))+\left (-2 a^3 b+4 a b^3\right ) \cos (4 (c+d x))+12 a^4 \sin (c+d x)-25 a^2 b^2 \sin (c+d x)+13 b^4 \sin (c+d x)+4 a^4 \sin (3 (c+d x))-9 a^2 b^2 \sin (3 (c+d x))+5 b^4 \sin (3 (c+d x))\right )}{24 (a-b)^2 (a+b)^2 d \sqrt {a+b \sin (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^4/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(16*a*(a^3 + a^2*b - 2*a*b^2 - 2*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x]
)/(a + b)] - 4*(4*a^4 - 9*a^2*b^2 + 5*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)] + Sec[c + d*x]^3*(-4*a^3*b + 10*a*b^3 + (-6*a^3*b + 14*a*b^3)*Cos[2*(c + d*x)] + (-2*a^3*b + 4
*a*b^3)*Cos[4*(c + d*x)] + 12*a^4*Sin[c + d*x] - 25*a^2*b^2*Sin[c + d*x] + 13*b^4*Sin[c + d*x] + 4*a^4*Sin[3*(
c + d*x)] - 9*a^2*b^2*Sin[3*(c + d*x)] + 5*b^4*Sin[3*(c + d*x)]))/(24*(a - b)^2*(a + b)^2*d*Sqrt[a + b*Sin[c +
d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1312\) vs. \(2(337)=674\).

Time = 2.02 (sec) , antiderivative size = 1313, normalized size of antiderivative = 4.51

method	result	size
default	\(\text {Expression too large to display}\)	\(1313\)

[In]

int(sec(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/cos(d*x+c)^5/(a+b*sin(d*x+c))^(3/2)/b/(a^4-2*a^2*b^2+b^4)*(-4*cos(
d*x+c)^4*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*a*b^2*(a^2-2*b^2)+cos(d*x+c)^2*(b*cos(d*x+c)^2*sin(d
*x+c)+a*cos(d*x+c)^2)^(1/2)*b*(4*a^4-9*a^2*b^2+5*b^4)*sin(d*x+c)+2*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^
(1/2)*b*(a^4-2*a^2*b^2+b^4)*sin(d*x+c)+cos(d*x+c)^2*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*(4*Ellipt
icE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d
*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^5-12*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),(
(a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c
)+b/(a+b))^(1/2)*a^3*b^2+8*EllipticE((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x
+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a*b^4-4*EllipticF((b
/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+
a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^4*b+3*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)
/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(
a+b))^(1/2)*a^3*b^2+9*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b
/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^2*b^3-3*EllipticF((b/(a
-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(
a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a*b^4-5*EllipticF((b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2),((a-b)/(a
+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b
))^(1/2)*b^5+a^3*b^2-a*b^4))/d

Fricas [C] (verification not implemented)

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

Time = 0.13 (sec) , antiderivative size = 571, normalized size of antiderivative = 1.96 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {\sqrt {2} {\left (8 \, a^{4} - 19 \, a^{2} b^{2} + 15 \, b^{4}\right )} \sqrt {i \, b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) + \sqrt {2} {\left (8 \, a^{4} - 19 \, a^{2} b^{2} + 15 \, b^{4}\right )} \sqrt {-i \, b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right ) - 12 \, \sqrt {2} {\left (-i \, a^{3} b + 2 i \, a b^{3}\right )} \sqrt {i \, b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) - 12 \, \sqrt {2} {\left (i \, a^{3} b - 2 i \, a b^{3}\right )} \sqrt {-i \, b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right ) - 6 \, {\left (2 \, a^{2} b^{2} - 2 \, b^{4} + {\left (a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b - a b^{3} + 2 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{36 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/36*(sqrt(2)*(8*a^4 - 19*a^2*b^2 + 15*b^4)*sqrt(I*b)*cos(d*x + c)^3*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/
b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b) + sqrt(2)*(8*a^4
- 19*a^2*b^2 + 15*b^4)*sqrt(-I*b)*cos(d*x + c)^3*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3
+ 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b) - 12*sqrt(2)*(-I*a^3*b + 2*I*a*b^3)*
sqrt(I*b)*cos(d*x + c)^3*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstras
sPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c
) - 2*I*a)/b)) - 12*sqrt(2)*(I*a^3*b - 2*I*a*b^3)*sqrt(-I*b)*cos(d*x + c)^3*weierstrassZeta(-4/3*(4*a^2 - 3*b^
2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*
a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) - 6*(2*a^2*b^2 - 2*b^4 + (a^2*b^2 - 5*b^4)
*cos(d*x + c)^2 - 2*(a^3*b - a*b^3 + 2*(a^3*b - 2*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a
))/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c)^3)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**4/(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**4/sqrt(a + b*sin(c + d*x)), x)

Maxima [F]

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

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/sqrt(b*sin(d*x + c) + a), x)

Giac [F]

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

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/sqrt(b*sin(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^(1/2)), x)