Integrand size = 23, antiderivative size = 207 \[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {\left (12 a^2-18 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{3/2} d}+\frac {\left (12 a^2+18 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{3/2} d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \tan (c+d x)}{4 d} \]
-1/32*(12*a^2-18*a*b+5*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a -b)^(3/2)/d+1/32*(12*a^2+18*a*b+5*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b )^(1/2))/(a+b)^(3/2)/d-1/16*sec(d*x+c)^2*(a*b-(6*a^2-5*b^2)*sin(d*x+c))*(a +b*sin(d*x+c))^(1/2)/(a^2-b^2)/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^(1/2)*t an(d*x+c)/d
Time = 1.01 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.08 \[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {-\sqrt {a-b} (a+b)^2 \left (12 a^2-18 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+(a-b)^2 \sqrt {a+b} \left (12 a^2+18 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+\frac {1}{2} \left (a^2-b^2\right ) \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \left (-2 a b-2 a b \cos (2 (c+d x))+\left (22 a^2-21 b^2\right ) \sin (c+d x)+6 a^2 \sin (3 (c+d x))-5 b^2 \sin (3 (c+d x))\right )}{32 \left (a^2-b^2\right )^2 d} \]
(-(Sqrt[a - b]*(a + b)^2*(12*a^2 - 18*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[ c + d*x]]/Sqrt[a - b]]) + (a - b)^2*Sqrt[a + b]*(12*a^2 + 18*a*b + 5*b^2)* ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + ((a^2 - b^2)*Sec[c + d*x]^ 4*Sqrt[a + b*Sin[c + d*x]]*(-2*a*b - 2*a*b*Cos[2*(c + d*x)] + (22*a^2 - 21 *b^2)*Sin[c + d*x] + 6*a^2*Sin[3*(c + d*x)] - 5*b^2*Sin[3*(c + d*x)]))/2)/ (32*(a^2 - b^2)^2*d)
Time = 0.49 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.32, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3147, 494, 27, 686, 27, 654, 25, 1480, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \sin (c+d x)}}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^5 \int \frac {\sqrt {a+b \sin (c+d x)}}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 494 |
| \(\displaystyle \frac {b^5 \left (\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {6 a+5 b \sin (c+d x)}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \left (\frac {\int \frac {6 a+5 b \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 686 |
| \(\displaystyle \frac {b^5 \left (\frac {-\frac {\int -\frac {a \left (12 a^2-13 b^2\right )+b \left (6 a^2-5 b^2\right ) \sin (c+d x)}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (a b^2-b \left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\int \frac {a \left (12 a^2-13 b^2\right )+b \left (6 a^2-5 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (a b^2-b \left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 654 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\int -\frac {b^2 \left (6 a^2-5 b^2\right ) \sin ^2(c+d x)+2 a \left (3 a^2-4 b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (a b^2-b \left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {-\frac {\int \frac {b^2 \left (6 a^2-5 b^2\right ) \sin ^2(c+d x)+2 a \left (3 a^2-4 b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (a b^2-b \left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {(a+b) \left (12 a^2-18 a b+5 b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a-b) \left (12 a^2+18 a b+5 b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (a b^2-b \left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\frac {(a-b) \left (12 a^2+18 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}-\frac {(a+b) \left (12 a^2-18 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b \sqrt {a-b}}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (a b^2-b \left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2}+\frac {\sin (c+d x) \sqrt {a+b \sin (c+d x)}}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
(b^5*((Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(4*b*(b^2 - b^2*Sin[c + d*x] ^2)^2) + ((-1/2*((a + b)*(12*a^2 - 18*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[ c + d*x]]/Sqrt[a - b]])/(Sqrt[a - b]*b) + ((a - b)*(12*a^2 + 18*a*b + 5*b^ 2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b*Sqrt[a + b]))/(2*b^ 2*(a^2 - b^2)) - (Sqrt[a + b*Sin[c + d*x]]*(a*b^2 - b*(6*a^2 - 5*b^2)*Sin[ c + d*x]))/(2*b^2*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)))/(8*b^2)))/d
3.5.78.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-x)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1)*(c*(2*p + 3) + d*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 0] && (Lt Q[n, 1] || (ILtQ[n + 2*p + 3, 0] && NeQ[n, 2])) && IntQuadraticQ[a, 0, b, c , d, n, p, x]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d* x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 2.16 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.14
| method | result | size |
| default | \(-\frac {2 b^{5} \left (-\frac {-\frac {b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, \left (6 a \sin \left (d x +c \right )-5 b \sin \left (d x +c \right )+8 a -7 b \right )}{4 \left (a -b \right ) \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {\left (12 a^{2}-18 a b +5 b^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \left (a -b \right ) \sqrt {-a +b}}}{16 b^{5}}+\frac {\frac {b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, \left (6 a \sin \left (d x +c \right )+5 b \sin \left (d x +c \right )-8 a -7 b \right )}{4 \left (a +b \right ) \left (b \sin \left (d x +c \right )-b \right )^{2}}-\frac {\left (12 a^{2}+18 a b +5 b^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \left (a +b \right )^{\frac {3}{2}}}}{16 b^{5}}\right )}{d}\) | \(237\) |
-2*b^5*(-1/16/b^5*(-1/4/(a-b)*b^2*(a+b*sin(d*x+c))^(1/2)*(6*a*sin(d*x+c)-5 *b*sin(d*x+c)+8*a-7*b)/(b*sin(d*x+c)+b)^2+1/4*(12*a^2-18*a*b+5*b^2)/(a-b)/ (-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)))+1/16/b^5*(1/4/(a +b)*b^2*(a+b*sin(d*x+c))^(1/2)*(6*a*sin(d*x+c)+5*b*sin(d*x+c)-8*a-7*b)/(b* sin(d*x+c)-b)^2-1/4*(12*a^2+18*a*b+5*b^2)/(a+b)^(3/2)*arctanh((a+b*sin(d*x +c))^(1/2)/(a+b)^(1/2))))/d
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (183) = 366\).
Time = 0.75 (sec) , antiderivative size = 2505, normalized size of antiderivative = 12.10 \[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
[1/256*((12*a^4 - 6*a^3*b - 19*a^2*b^2 + 8*a*b^3 + 5*b^4)*sqrt(a + b)*cos( d*x + c)^4*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 320*a^2*b^2 + 2 56*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 + 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*( 16*a^3 + 24*a^2*b + 20*a*b^2 + 8*b^3 - (10*a*b^2 + 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*s in(d*x + c) + a)*sqrt(a + b) + 4*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b ^4 - (8*a*b^3 + 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*c os(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - (12*a^4 + 6*a^ 3*b - 19*a^2*b^2 - 8*a*b^3 + 5*b^4)*sqrt(a - b)*cos(d*x + c)^4*log((b^4*co s(d*x + c)^4 + 128*a^4 - 256*a^3*b + 320*a^2*b^2 - 256*a*b^3 + 72*b^4 - 8* (20*a^2*b^2 - 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 - 24*a^2*b + 20 *a*b^2 - 8*b^3 - (10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b + 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt( a - b) + 4*(64*a^3*b - 112*a^2*b^2 + 64*a*b^3 - 14*b^4 - (8*a*b^3 - 7*b^4) *cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos (d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*((a^3*b - a*b^3)*cos(d*x + c)^2 - (4*a^4 - 8*a^2*b^2 + 4*b^4 + (6*a^4 - 11*a^2*b^2 + 5*b^4)*cos(d*x + c)^2) *sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d* x + c)^4), -1/256*(2*(12*a^4 - 6*a^3*b - 19*a^2*b^2 + 8*a*b^3 + 5*b^4)*sqr t(-a - b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(...
\[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int \sqrt {a + b \sin {\left (c + d x \right )}} \sec ^{5}{\left (c + d x \right )}\, dx \]
Exception generated. \[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for m ore detail
Time = 0.32 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.73 \[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {b^{5} {\left (\frac {{\left (12 \, a^{2} - 18 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a + b}}\right )}{{\left (a b^{5} - b^{6}\right )} \sqrt {-a + b}} - \frac {{\left (12 \, a^{2} + 18 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a - b}}\right )}{{\left (a b^{5} + b^{6}\right )} \sqrt {-a - b}} - \frac {2 \, {\left (6 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 18 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 18 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 6 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{5} - 5 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} b^{2} + 14 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a b^{2} - 23 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} b^{2} + 14 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{3} b^{2} + 9 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{4} - 8 \, \sqrt {b \sin \left (d x + c\right ) + a} a b^{4}\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} {\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (b \sin \left (d x + c\right ) + a\right )} a + a^{2} - b^{2}\right )}^{2}}\right )}}{32 \, d} \]
1/32*b^5*((12*a^2 - 18*a*b + 5*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(- a + b))/((a*b^5 - b^6)*sqrt(-a + b)) - (12*a^2 + 18*a*b + 5*b^2)*arctan(sq rt(b*sin(d*x + c) + a)/sqrt(-a - b))/((a*b^5 + b^6)*sqrt(-a - b)) - 2*(6*( b*sin(d*x + c) + a)^(7/2)*a^2 - 18*(b*sin(d*x + c) + a)^(5/2)*a^3 + 18*(b* sin(d*x + c) + a)^(3/2)*a^4 - 6*sqrt(b*sin(d*x + c) + a)*a^5 - 5*(b*sin(d* x + c) + a)^(7/2)*b^2 + 14*(b*sin(d*x + c) + a)^(5/2)*a*b^2 - 23*(b*sin(d* x + c) + a)^(3/2)*a^2*b^2 + 14*sqrt(b*sin(d*x + c) + a)*a^3*b^2 + 9*(b*sin (d*x + c) + a)^(3/2)*b^4 - 8*sqrt(b*sin(d*x + c) + a)*a*b^4)/((a^2*b^4 - b ^6)*((b*sin(d*x + c) + a)^2 - 2*(b*sin(d*x + c) + a)*a + a^2 - b^2)^2))/d
Timed out. \[ \int \sec ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int \frac {\sqrt {a+b\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^5} \,d x \]