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

Optimal. Leaf size=291 \[ -\frac {\sec ^3(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-5 b^2\right )-4 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^2}+\frac {\left (4 a^2-5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.44, antiderivative size = 291, 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 = {2696, 2866, 2752, 2663, 2661, 2655, 2653} \[ -\frac {\sec ^3(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-5 b^2\right )-4 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^2}+\frac {\left (4 a^2-5 b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(b - a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(3*(a^2 - b^2)*d) - (2*a*(a^2 - 2*b^2)*Ellipti
cE[(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 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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[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 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 \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 {\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 ) 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}}}{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*}

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Mathematica [A]  time = 4.14, size = 306, normalized size = 1.05 \[ \frac {-4 \left (4 a^4-9 a^2 b^2+5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+16 a \left (a^3+a^2 b-2 a b^2-2 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\sec ^3(c+d x) \left (12 a^4 \sin (c+d x)+4 a^4 \sin (3 (c+d x))+\left (14 a b^3-6 a^3 b\right ) \cos (2 (c+d x))+\left (4 a b^3-2 a^3 b\right ) \cos (4 (c+d x))-4 a^3 b-25 a^2 b^2 \sin (c+d x)-9 a^2 b^2 \sin (3 (c+d x))+10 a b^3+13 b^4 \sin (c+d x)+5 b^4 \sin (3 (c+d x))\right )}{24 d (a-b)^2 (a+b)^2 \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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]])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (d x + c\right )^{4}}{\sqrt {b \sin \left (d x + c\right ) + a}}, x\right ) \]

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

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

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

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maple [B]  time = 2.26, size = 1314, normalized size = 4.52 \[ \text {result too large to display} \]

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*(-(-a-b*sin(d*x+c))*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*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*a*b^2*(a^2-2*b^2)+cos(d*x+c)^2*(cos(d*x+c)^2*sin(d*x
+c)*b+cos(d*x+c)^2*a)^(1/2)*b*(4*a^4-9*a^2*b^2+5*b^4)*sin(d*x+c)+2*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^
(1/2)*b*(a^4-2*a^2*b^2+b^4)*sin(d*x+c)-cos(d*x+c)^2*(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)+b/(a+b))^(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)+1/(a-b)*a)^(1/2)*a^4*b-3*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(
a+b)*sin(d*x+c)+b/(a+b))^(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)*si
n(d*x+c)+1/(a-b)*a)^(1/2)*a^3*b^2-9*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*El
lipticF((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)+1/(a-b)*a)^(1/2)*a^2*b^3
+3*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(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)+1/(a-b)*a)^(1/2)*a*b^4+5*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1
/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(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)+1/(a-b)*a)^(1/2)*b^5-4*(-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)+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))*a^
5+12*(-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)+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))*a^3*b^2-8*(-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)+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))*a*b^4-a^3*b^2+a*b^4))/d

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

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

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