\(\int \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) [482]

Optimal result

Integrand size = 23, antiderivative size = 248 \[ \int \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {\left (4 a^2-3 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{6 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{3 d} \]

[Out]

-1/6*sec(d*x+c)*(a*b-(4*a^2-3*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)/d+1/6*(4*a^2-3*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)*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)/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-2/3*a*(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)/d/(a+b*sin(d*x+c))^(1/2)+1/3*sec(d*x+c)^2*(a+b*sin(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 248, 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 = {2769, 2945, 2831, 2742, 2740, 2734, 2732} \[ \int \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )}-\frac {\left (4 a^2-3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \sin (c+d x)}}+\frac {\tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 d} \]

[In]

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

[Out]

-1/6*((4*a^2 - 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)*d*Sq
rt[(a + b*Sin[c + d*x])/(a + b)]) + (2*a*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x]
)/(a + b)])/(3*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*b - (4*a^2 - 3*b^2)*Sin
[c + d*x]))/(6*(a^2 - b^2)*d) + (Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]]*Tan[c + d*x])/(3*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 2769

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

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 ^2(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{3 d}-\frac {1}{3} \int \frac {\sec ^2(c+d x) \left (-2 a-\frac {3}{2} b \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx \\ & = -\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{3 d}+\frac {\int \frac {-\frac {a b^2}{4}-\frac {1}{4} b \left (4 a^2-3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )} \\ & = -\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} a \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx-\frac {\left (4 a^2-3 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{12 \left (a^2-b^2\right )} \\ & = -\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{3 d}-\frac {\left (\left (4 a^2-3 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}{12 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (a \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}{3 \sqrt {a+b \sin (c+d x)}} \\ & = -\frac {\left (4 a^2-3 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)}}{6 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \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}}}{3 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.09 \[ \int \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {\left (4 a^3+4 a^2 b-3 a b^2-3 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-4 a \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+\frac {1}{8} \sec ^3(c+d x) \left (8 a^2 b-11 b^3+\left (-12 a^2 b+8 b^3\right ) \cos (2 (c+d x))+\left (-4 a^2 b+3 b^3\right ) \cos (4 (c+d x))+24 a^3 \sin (c+d x)-24 a b^2 \sin (c+d x)+8 a^3 \sin (3 (c+d x))-8 a b^2 \sin (3 (c+d x))\right )}{6 (a-b) (a+b) d \sqrt {a+b \sin (c+d x)}} \]

[In]

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

[Out]

((4*a^3 + 4*a^2*b - 3*a*b^2 - 3*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])
/(a + b)] - 4*a*(a^2 - b^2)*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*(8*a^2*b - 11*b^3 + (-12*a^2*b + 8*b^3)*Cos[2*(c + d*x)] + (-4*a^2*b + 3*b^3)*Cos[4*(c + d*
x)] + 24*a^3*Sin[c + d*x] - 24*a*b^2*Sin[c + d*x] + 8*a^3*Sin[3*(c + d*x)] - 8*a*b^2*Sin[3*(c + d*x)]))/8)/(6*
(a - b)*(a + b)*d*Sqrt[a + b*Sin[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1200\) vs. \(2(294)=588\).

Time = 2.46 (sec) , antiderivative size = 1201, normalized size of antiderivative = 4.84

method	result	size
default	\(\text {Expression too large to display}\)	\(1201\)

[In]

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

[Out]

1/6*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*(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)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b
))^(1/2)*a^3*b*cos(d*x+c)^2-3*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)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*a^2*b^2*cos(d*x+c
)^2-4*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)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*a*b^3*cos(d*x+c)^2+3*EllipticF((b/(a-b)*s
in(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)+b/(a+b)
)^(1/2)*(b/(a-b)*sin(d*x+c)+a/(a-b))^(1/2)*b^4*cos(d*x+c)^2-4*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*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)*a^4*cos(d*x+c)^2+7*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*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)*a^2*b^2*cos(d
*x+c)^2-3*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*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^4*cos(d*x+c)^2+4*a^2*b^2*cos(d*x+c)
^4-3*b^4*cos(d*x+c)^4-4*a^3*b*cos(d*x+c)^2*sin(d*x+c)+4*a*b^3*cos(d*x+c)^2*sin(d*x+c)-a^2*b^2*cos(d*x+c)^2+b^4
*cos(d*x+c)^2-2*a^3*b*sin(d*x+c)+2*a*b^3*sin(d*x+c)-2*a^2*b^2+2*b^4)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+sin(
d*x+c)))^(1/2)/(a+b)/(a-b)/(1+sin(d*x+c))/(sin(d*x+c)-1)/b/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/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 = 526, normalized size of antiderivative = 2.12 \[ \int \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (8 \, a^{3} - 9 \, a b^{2}\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^{3} - 9 \, a b^{2}\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 ) + 3 \, \sqrt {2} {\left (4 i \, a^{2} b - 3 i \, 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 ) + 3 \, \sqrt {2} {\left (-4 i \, a^{2} b + 3 i \, 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 (a b^{2} \cos \left (d x + c\right )^{2} - {\left (2 \, a^{2} b - 2 \, b^{3} + {\left (4 \, a^{2} b - 3 \, 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^{2} b - b^{3}\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^3 - 9*a*b^2)*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^3 - 9*a*b^2)*s
qrt(-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) + 3*sqrt(2)*(4*I*a^2*b - 3*I*b^3)*sqrt(I*b)*cos(d*x + c)^3*w
eierstrassZeta(-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)) + 3*sqrt(2)
*(-4*I*a^2*b + 3*I*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*(a*b^2*cos(d*x + c)^2 - (2*a^2*b - 2*b^3 + (4*a^2*b - 3*b^3)*cos
(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^2*b - b^3)*d*cos(d*x + c)^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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