3.6.26 \(\int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [526]

Optimal. Leaf size=359 \[ \frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^4-15 a^2 b^2-21 b^4\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{6 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \left (a^2-3 b^2\right ) F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d} \]

[Out]

2*b*sec(d*x+c)^3/(a^2-b^2)/d/(a+b*sin(d*x+c))^(1/2)-1/3*sec(d*x+c)^3*(8*a*b-(a^2+7*b^2)*sin(d*x+c))*(a+b*sin(d
*x+c))^(1/2)/(a^2-b^2)^2/d-1/6*sec(d*x+c)*(a*b*(a^2-33*b^2)-(4*a^4-15*a^2*b^2-21*b^4)*sin(d*x+c))*(a+b*sin(d*x
+c))^(1/2)/(a^2-b^2)^3/d+1/6*(4*a^4-15*a^2*b^2-21*b^4)*(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)^3/d/((a+b
*sin(d*x+c))/(a+b))^(1/2)-2/3*a*(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)*Elli
pticF(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)^2/d/(a+b*sin
(d*x+c))^(1/2)

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Rubi [A]
time = 0.40, antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2773, 2945, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac {2 b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}+\frac {2 a \left (a^2-3 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 )}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^3}-\frac {\left (4 a^4-15 a^2 b^2-21 b^4\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 )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(2*b*Sec[c + d*x]^3)/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((4*a^4 - 15*a^2*b^2 - 21*b^4)*EllipticE[(c -
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(6*(a^2 - b^2)^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)])
+ (2*a*(a^2 - 3*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*(a^2
- b^2)^2*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(8*a*b - (a^2 + 7*b^2)*Sin[c
+ d*x]))/(3*(a^2 - b^2)^2*d) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*b*(a^2 - 33*b^2) - (4*a^4 - 15*a^2*b^
2 - 21*b^4)*Sin[c + d*x]))/(6*(a^2 - b^2)^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 2773

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - b^2)*(m + 1))), x] + Dist[1/((a^2 - b^2)*(m
+ 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -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*} \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=\frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {2 \int \frac {\sec ^4(c+d x) \left (-\frac {a}{2}+\frac {7}{2} b \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {2 \int \frac {\sec ^2(c+d x) \left (a \left (a^2-3 b^2\right )+\frac {3}{4} b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac {2 \int \frac {\frac {1}{8} a b^2 \left (a^2-33 b^2\right )+\frac {1}{8} b \left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^3}\\ &=\frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac {\left (a \left (a^2-3 b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}-\frac {\left (4 a^4-15 a^2 b^2-21 b^4\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{12 \left (a^2-b^2\right )^3}\\ &=\frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac {\left (\left (4 a^4-15 a^2 b^2-21 b^4\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 )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (a \left (a^2-3 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}{3 \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^4-15 a^2 b^2-21 b^4\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 )^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \left (a^2-3 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}}}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]
time = 3.03, size = 348, normalized size = 0.97 \begin {gather*} \frac {\left (4 a^5+4 a^4 b-15 a^3 b^2-15 a^2 b^3-21 a b^4-21 b^5\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^4-4 a^2 b^2+3 b^4\right ) F\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 (-24 a^4 b+101 a^2 b^3+19 b^5+\left (-12 a^4 b+84 a^2 b^3+56 b^5\right ) \cos (2 (c+d x))+\left (-4 a^4 b+15 a^2 b^3+21 b^5\right ) \cos (4 (c+d x))+24 a^5 \sin (c+d x)-64 a^3 b^2 \sin (c+d x)+40 a b^4 \sin (c+d x)+8 a^5 \sin (3 (c+d x))-32 a^3 b^2 \sin (3 (c+d x))+24 a b^4 \sin (3 (c+d x))\right )}{6 (a-b)^3 (a+b)^3 d \sqrt {a+b \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

((4*a^5 + 4*a^4*b - 15*a^3*b^2 - 15*a^2*b^3 - 21*a*b^4 - 21*b^5)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b
)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - 4*a*(a^4 - 4*a^2*b^2 + 3*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*(-24*a^4*b + 101*a^2*b^3 + 19*b^5 + (-12*a^4*b +
84*a^2*b^3 + 56*b^5)*Cos[2*(c + d*x)] + (-4*a^4*b + 15*a^2*b^3 + 21*b^5)*Cos[4*(c + d*x)] + 24*a^5*Sin[c + d*x
] - 64*a^3*b^2*Sin[c + d*x] + 40*a*b^4*Sin[c + d*x] + 8*a^5*Sin[3*(c + d*x)] - 32*a^3*b^2*Sin[3*(c + d*x)] + 2
4*a*b^4*Sin[3*(c + d*x)]))/8)/(6*(a - b)^3*(a + b)^3*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1645\) vs. \(2(403)=806\).
time = 15.54, size = 1646, normalized size = 4.58

method	result	size
default	\(\text {Expression too large to display}\)	\(1646\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[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^6-3*a^4*b^2+3*a^2*b^4-b^6
)*(-2*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*b^2*(a^4-2*a^2*b^2+b^4)-cos(d*x+c)^4*(b*cos(d*x+c)^2*si
n(d*x+c)+a*cos(d*x+c)^2)^(1/2)*b^2*(4*a^4-15*a^2*b^2-21*b^4)+4*cos(d*x+c)^2*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d
*x+c)^2)^(1/2)*a*b*(a^4-4*a^2*b^2+3*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)*a*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*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/(a-b)*sin(d*x+c)-b/
(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*a^6-19*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/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(
a-b)*a)^(1/2)*a^4*b^2-6*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/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*a^2*b^4+21*Elliptic
E((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/(a-b)*sin(
d*x+c)-b/(a-b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*b^6-4*(-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^5*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)*Elliptic
F((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^2+16*(-
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^3*b^3+18*(-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^2*b^4-12*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^
(1/2)*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^5-21*(-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^6+a^4*b^2+6*a^2*b^4-7*b^6))/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(b*sin(d*x + c) + a)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.22, size = 878, normalized size = 2.45 \begin {gather*} \frac {{\left (\sqrt {2} {\left (8 \, a^{5} b - 33 \, a^{3} b^{3} + 57 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \sqrt {2} {\left (8 \, a^{6} - 33 \, a^{4} b^{2} + 57 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {i \, b} {\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 ) + {\left (\sqrt {2} {\left (8 \, a^{5} b - 33 \, a^{3} b^{3} + 57 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \sqrt {2} {\left (8 \, a^{6} - 33 \, a^{4} b^{2} + 57 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-i \, b} {\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 \, {\left (\sqrt {2} {\left (4 i \, a^{4} b^{2} - 15 i \, a^{2} b^{4} - 21 i \, b^{6}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \sqrt {2} {\left (4 i \, a^{5} b - 15 i \, a^{3} b^{3} - 21 i \, a b^{5}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {i \, b} {\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 \, {\left (\sqrt {2} {\left (-4 i \, a^{4} b^{2} + 15 i \, a^{2} b^{4} + 21 i \, b^{6}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \sqrt {2} {\left (-4 i \, a^{5} b + 15 i \, a^{3} b^{3} + 21 i \, a b^{5}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-i \, b} {\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^{4} b^{2} - 4 \, a^{2} b^{4} + 2 \, b^{6} + {\left (4 \, a^{4} b^{2} - 15 \, a^{2} b^{4} - 21 \, b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{4} b^{2} + 6 \, a^{2} b^{4} - 7 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5} + 2 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\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 ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/36*((sqrt(2)*(8*a^5*b - 33*a^3*b^3 + 57*a*b^5)*cos(d*x + c)^3*sin(d*x + c) + sqrt(2)*(8*a^6 - 33*a^4*b^2 + 5
7*a^2*b^4)*cos(d*x + c)^3)*sqrt(I*b)*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^5*b - 33*a^3*b^3 + 57*a*b^5)*cos(
d*x + c)^3*sin(d*x + c) + sqrt(2)*(8*a^6 - 33*a^4*b^2 + 57*a^2*b^4)*cos(d*x + c)^3)*sqrt(-I*b)*weierstrassPInv
erse(-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^4*b^2 - 15*I*a^2*b^4 - 21*I*b^6)*cos(d*x + c)^3*sin(d*x + c) + sqrt(2)*(4*I*a^5*
b - 15*I*a^3*b^3 - 21*I*a*b^5)*cos(d*x + c)^3)*sqrt(I*b)*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)) + 3*(sqrt(2)*(-4*I*a^4*b^2 + 15*I*a^2*b^4 + 21*I*b^6)*cos(d*x +
c)^3*sin(d*x + c) + sqrt(2)*(-4*I*a^5*b + 15*I*a^3*b^3 + 21*I*a*b^5)*cos(d*x + c)^3)*sqrt(-I*b)*weierstrassZe
ta(-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^4*b^2 - 4*a^
2*b^4 + 2*b^6 + (4*a^4*b^2 - 15*a^2*b^4 - 21*b^6)*cos(d*x + c)^4 - (a^4*b^2 + 6*a^2*b^4 - 7*b^6)*cos(d*x + c)^
2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5 + 2*(a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*
x + c) + a))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^3*sin(d*x + c) + (a^7*b - 3*a^5*b^3 + 3*a
^3*b^5 - a*b^7)*d*cos(d*x + c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**4/(a + b*sin(c + d*x))**(3/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^(3/2)), x)

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