Integrand size = 23, antiderivative size = 230 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=-\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{5/2} d}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{5/2} d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d} \]
-3/32*(4*a^2-10*a*b+7*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a- b)^(5/2)/d+3/32*(4*a^2+10*a*b+7*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^ (1/2))/(a+b)^(5/2)/d-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^(1 /2)/(a^2-b^2)/d-1/16*sec(d*x+c)^2*(b*(a^2-7*b^2)-6*a*(a^2-2*b^2)*sin(d*x+c ))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^2/d
Time = 1.25 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {-3 (a+b)^{5/2} \left (4 a^2-10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+\sqrt {a-b} \left (3 (a-b)^2 \left (4 a^2+10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+\sqrt {a+b} \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \left (-9 a^2 b+15 b^3+\left (-a^2 b+7 b^3\right ) \cos (2 (c+d x))+a \left (11 a^2-14 b^2\right ) \sin (c+d x)+3 \left (a^3-2 a b^2\right ) \sin (3 (c+d x))\right )\right )}{32 \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )^2 d} \]
(-3*(a + b)^(5/2)*(4*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x] ]/Sqrt[a - b]] + Sqrt[a - b]*(3*(a - b)^2*(4*a^2 + 10*a*b + 7*b^2)*ArcTanh [Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + Sqrt[a + b]*Sec[c + d*x]^4*Sqrt[a + b*Sin[c + d*x]]*(-9*a^2*b + 15*b^3 + (-(a^2*b) + 7*b^3)*Cos[2*(c + d*x) ] + a*(11*a^2 - 14*b^2)*Sin[c + d*x] + 3*(a^3 - 2*a*b^2)*Sin[3*(c + d*x)]) ))/(32*Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)^2*d)
Time = 0.55 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.37, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3147, 496, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^5 \sqrt {a+b \sin (c+d x)}}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^5 \int \frac {1}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {b^5 \left (\frac {\int \frac {6 a^2+5 b \sin (c+d x) a-7 b^2}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \left (\frac {\int \frac {6 a^2+5 b \sin (c+d x) a-7 b^2}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 686 |
| \(\displaystyle \frac {b^5 \left (\frac {-\frac {\int -\frac {3 \left (4 a^4-9 b^2 a^2+2 b \left (a^2-2 b^2\right ) \sin (c+d x) a+7 b^4\right )}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2 \left (a^2-7 b^2\right )-6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {3 \int \frac {4 a^4-9 b^2 a^2+2 b \left (a^2-2 b^2\right ) \sin (c+d x) a+7 b^4}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2 \left (a^2-7 b^2\right )-6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 654 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {3 \int -\frac {2 a^4-5 b^2 a^2+2 b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x) a+7 b^4}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2 \left (a^2-7 b^2\right )-6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {-\frac {3 \int \frac {2 a^4-5 b^2 a^2+2 b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x) a+7 b^4}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2 \left (a^2-7 b^2\right )-6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {3 \left (\frac {(a+b)^2 \left (4 a^2-10 a b+7 b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a-b)^2 \left (4 a^2+10 a b+7 b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2 \left (a^2-7 b^2\right )-6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {3 \left (\frac {(a-b)^2 \left (4 a^2+10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}-\frac {(a+b)^2 \left (4 a^2-10 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b \sqrt {a-b}}\right )}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2 \left (a^2-7 b^2\right )-6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{8 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{4 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
(b^5*(-1/4*(Sqrt[a + b*Sin[c + d*x]]*(b^2 - a*b*Sin[c + d*x]))/(b^2*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)^2) + ((3*(-1/2*((a + b)^2*(4*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(Sqrt[a - b]*b) + ((a - b)^2*(4*a^2 + 10*a*b + 7*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]]*(b^2*(a^2 - 7*b^2) - 6*a*b*(a^2 - 2*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*(a^2 - b^2))))/d
3.6.11.3.1 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[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[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 = 1.57 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.17
| method | result | size |
| default | \(-\frac {2 b^{5} \left (-\frac {-\frac {b^{2} \left (6 a \sin \left (d x +c \right )-9 b \sin \left (d x +c \right )+8 a -11 b \right ) \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a^{2}-2 a b +b^{2}\right ) \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 \left (4 a^{2}-10 a b +7 b^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}}{16 b^{5}}+\frac {\frac {b^{2} \left (6 a \sin \left (d x +c \right )+9 b \sin \left (d x +c \right )-8 a -11 b \right ) \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a^{2}+2 a b +b^{2}\right ) \left (b \sin \left (d x +c \right )-b \right )^{2}}-\frac {3 \left (4 a^{2}+10 a b +7 b^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}}}{16 b^{5}}\right )}{d}\) | \(270\) |
-2*b^5*(-1/16/b^5*(-1/4/(a^2-2*a*b+b^2)*b^2*(6*a*sin(d*x+c)-9*b*sin(d*x+c) +8*a-11*b)*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)^2+3/4*(4*a^2-10*a*b+7*b ^2)/(a^2-2*a*b+b^2)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2 )))+1/16/b^5*(1/4/(a^2+2*a*b+b^2)*b^2*(6*a*sin(d*x+c)+9*b*sin(d*x+c)-8*a-1 1*b)*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)^2-3/4*(4*a^2+10*a*b+7*b^2)/(a ^2+2*a*b+b^2)/(a+b)^(1/2)*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. 543 vs. \(2 (208) = 416\).
Time = 0.87 (sec) , antiderivative size = 2721, normalized size of antiderivative = 11.83 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\text {Too large to display} \]
[1/256*(3*(4*a^5 - 2*a^4*b - 11*a^3*b^2 + 5*a^2*b^3 + 11*a*b^4 - 7*b^5)*sq rt(a + b)*cos(d*x + c)^4*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 3 20*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)*co s(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 + 6 4*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)) + 3*(4*a^5 + 2*a^4*b - 11*a^3*b^2 - 5*a^2*b^3 + 11*a*b^4 + 7*b^5)*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 - 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))*sq rt(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*(4*a^ 4*b - 8*a^2*b^3 + 4*b^5 + (a^4*b - 8*a^2*b^3 + 7*b^5)*cos(d*x + c)^2 - 2*( 2*a^5 - 4*a^3*b^2 + 2*a*b^4 + 3*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cos(d*x + c)^2 )*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4), -1/256*(6*(4*a^5 - 2*a^4*b - 11*a^3*b^2 + 5*a^2...
\[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]
Exception generated. \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (208) = 416\).
Time = 0.39 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.82 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {b^{5} {\left (\frac {3 \, {\left (4 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a + b}}\right )}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} \sqrt {-a + b}} - \frac {3 \, {\left (4 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a - b}}\right )}{{\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} \sqrt {-a - b}} - \frac {2 \, {\left (6 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - 18 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} + 18 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} - 6 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{6} - 12 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a b^{2} + 35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} b^{2} - 44 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} b^{2} + 21 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{4} b^{2} + 7 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} b^{4} + 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a b^{4} - 4 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{2} b^{4} - 11 \, \sqrt {b \sin \left (d x + c\right ) + a} b^{6}\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} {\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (b \sin \left (d x + c\right ) + a\right )} a + a^{2} - b^{2}\right )}^{2}}\right )}}{32 \, d} \]
1/32*b^5*(3*(4*a^2 - 10*a*b + 7*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt( -a + b))/((a^2*b^5 - 2*a*b^6 + b^7)*sqrt(-a + b)) - 3*(4*a^2 + 10*a*b + 7* b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a - b))/((a^2*b^5 + 2*a*b^6 + b ^7)*sqrt(-a - b)) - 2*(6*(b*sin(d*x + c) + a)^(7/2)*a^3 - 18*(b*sin(d*x + c) + a)^(5/2)*a^4 + 18*(b*sin(d*x + c) + a)^(3/2)*a^5 - 6*sqrt(b*sin(d*x + c) + a)*a^6 - 12*(b*sin(d*x + c) + a)^(7/2)*a*b^2 + 35*(b*sin(d*x + c) + a)^(5/2)*a^2*b^2 - 44*(b*sin(d*x + c) + a)^(3/2)*a^3*b^2 + 21*sqrt(b*sin(d *x + c) + a)*a^4*b^2 + 7*(b*sin(d*x + c) + a)^(5/2)*b^4 + 2*(b*sin(d*x + c ) + a)^(3/2)*a*b^4 - 4*sqrt(b*sin(d*x + c) + a)*a^2*b^4 - 11*sqrt(b*sin(d* x + c) + a)*b^6)/((a^4*b^4 - 2*a^2*b^6 + b^8)*((b*sin(d*x + c) + a)^2 - 2* (b*sin(d*x + c) + a)*a + a^2 - b^2)^2))/d
Timed out. \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^5\,\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]