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

Optimal. Leaf size=248 \[ -\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )}-\frac {\left (4 a^2-3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \sin (c+d x)}}+\frac {\tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 d} \]

[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)/d/(a^2-b^2)+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]  time = 0.37, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2690, 2866, 2752, 2663, 2661, 2655, 2653} \[ -\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )}-\frac {\left (4 a^2-3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \sin (c+d x)}}+\frac {\tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*a^2 - 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(6*(a^2 - b^2)*d*Sqrt
[(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 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2690

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*Cos
[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 2752

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 2866

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*} \int \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \, dx &=\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 F\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]  time = 3.36, size = 270, normalized size = 1.09 \[ \frac {-4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\left (4 a^3+4 a^2 b-3 a b^2-3 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\frac {1}{8} \sec ^3(c+d x) \left (24 a^3 \sin (c+d x)+8 a^3 \sin (3 (c+d x))+\left (8 b^3-12 a^2 b\right ) \cos (2 (c+d x))+\left (3 b^3-4 a^2 b\right ) \cos (4 (c+d x))+8 a^2 b-24 a b^2 \sin (c+d x)-8 a b^2 \sin (3 (c+d x))-11 b^3\right )}{6 d (a-b) (a+b) \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[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]])

________________________________________________________________________________________

fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [B]  time = 0.85, size = 1259, normalized size = 5.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-4*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*a*b*(a^2-b^2)*sin(d*x+c)*cos(d*x+c)^2-2*(cos(d*x+c)^2
*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*a*b*(a^2-b^2)*sin(d*x+c)+(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*
b^2*(4*a^2-3*b^2)*cos(d*x+c)^4+(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*(4*(-b/(a-b)*sin(d*x+c)-b/(a-b
))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/
2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^3*b-3*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b
)*a)^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(
1/2)*a^2*b^2-4*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticF((b/(a-b)*sin
(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a*b^3+3*(-b/(a-b)*sin(d*x+c)
-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+
b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*b^4-4*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1
/(a-b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+
b))^(1/2)*a^4+7*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticE((b/(a-b)*si
n(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^2*b^2-3*(-b/(a-b)*sin(d*x
+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/
(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*b^4-a^2*b^2+b^4)*cos(d*x+c)^2-2*(cos(d*x+c)^2*sin(d*x+c)*b+c
os(d*x+c)^2*a)^(1/2)*a^2*b^2+2*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*b^4)/(-(a+b*sin(d*x+c))*(sin(d
*x+c)-1)*(1+sin(d*x+c)))^(1/2)/(a+b)/(sin(d*x+c)-1)/(a-b)/(1+sin(d*x+c))/b/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+b\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sin {\left (c + d x \right )}} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________