Integrand size = 24, antiderivative size = 249 \[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {b^{5/4} \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^3 d}+\frac {\left (3 a^2-6 a b+35 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 (a-b)^3 d}-\frac {b^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^3 d}+\frac {1}{16 (a-b) d (1-\sin (c+d x))^2}+\frac {3 a-11 b}{16 (a-b)^2 d (1-\sin (c+d x))}-\frac {1}{16 (a-b) d (1+\sin (c+d x))^2}-\frac {3 a-11 b}{16 (a-b)^2 d (1+\sin (c+d x))} \]
1/8*(3*a^2-6*a*b+35*b^2)*arctanh(sin(d*x+c))/(a-b)^3/d+1/16/(a-b)/d/(1-sin (d*x+c))^2+1/16*(3*a-11*b)/(a-b)^2/d/(1-sin(d*x+c))-1/16/(a-b)/d/(1+sin(d* x+c))^2+1/16*(-3*a+11*b)/(a-b)^2/d/(1+sin(d*x+c))-1/2*b^(5/4)*arctanh(b^(1 /4)*sin(d*x+c)/a^(1/4))/a^(3/4)/d/(a^(1/2)-b^(1/2))^3+1/2*b^(5/4)*arctan(b ^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4)/d/(a^(1/2)+b^(1/2))^3
Result contains complex when optimal does not.
Time = 5.87 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {2 \left (3 a^2-6 a b+35 b^2\right ) \text {arctanh}(\sin (c+d x))}{(a-b)^3}+\frac {4 b^{5/4} \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^3}+\frac {4 i b^{5/4} \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^3}-\frac {4 i b^{5/4} \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^3}-\frac {4 b^{5/4} \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^3}+\frac {1}{(a-b) (-1+\sin (c+d x))^2}+\frac {-3 a+11 b}{(a-b)^2 (-1+\sin (c+d x))}-\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {-3 a+11 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \]
((2*(3*a^2 - 6*a*b + 35*b^2)*ArcTanh[Sin[c + d*x]])/(a - b)^3 + (4*b^(5/4) *Log[a^(1/4) - b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] - Sqrt[b])^3) + (( 4*I)*b^(5/4)*Log[a^(1/4) - I*b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] + Sq rt[b])^3) - ((4*I)*b^(5/4)*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*x]])/(a^(3/4) *(Sqrt[a] + Sqrt[b])^3) - (4*b^(5/4)*Log[a^(1/4) + b^(1/4)*Sin[c + d*x]])/ (a^(3/4)*(Sqrt[a] - Sqrt[b])^3) + 1/((a - b)*(-1 + Sin[c + d*x])^2) + (-3* a + 11*b)/((a - b)^2*(-1 + Sin[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x])^ 2) + (-3*a + 11*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)
Time = 0.51 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3702, 1485, 2009}
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.
\(\displaystyle \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x)^5 \left (a-b \sin (c+d x)^4\right )}dx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\sin ^2(c+d x)\right )^3 \left (a-b \sin ^4(c+d x)\right )}d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 1485 |
\(\displaystyle \frac {\int \left (\frac {\left (-\left ((a+3 b) \sin ^2(c+d x)\right )-3 a-b\right ) b^2}{(a-b)^3 \left (a-b \sin ^4(c+d x)\right )}+\frac {-3 a^2+6 b a-35 b^2}{8 (a-b)^3 \left (\sin ^2(c+d x)-1\right )}+\frac {3 a-11 b}{16 (a-b)^2 (\sin (c+d x)-1)^2}+\frac {3 a-11 b}{16 (a-b)^2 (\sin (c+d x)+1)^2}-\frac {1}{8 (a-b) (\sin (c+d x)-1)^3}+\frac {1}{8 (a-b) (\sin (c+d x)+1)^3}\right )d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^{5/4} \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^3}-\frac {b^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^3}+\frac {\left (3 a^2-6 a b+35 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 (a-b)^3}+\frac {3 a-11 b}{16 (a-b)^2 (1-\sin (c+d x))}-\frac {3 a-11 b}{16 (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 (a-b) (1-\sin (c+d x))^2}-\frac {1}{16 (a-b) (\sin (c+d x)+1)^2}}{d}\) |
((b^(5/4)*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sq rt[b])^3) + ((3*a^2 - 6*a*b + 35*b^2)*ArcTanh[Sin[c + d*x]])/(8*(a - b)^3) - (b^(5/4)*ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] - Sqrt[b])^3) + 1/(16*(a - b)*(1 - Sin[c + d*x])^2) + (3*a - 11*b)/(16*(a - b)^2*(1 - Sin[c + d*x])) - 1/(16*(a - b)*(1 + Sin[c + d*x])^2) - (3*a - 1 1*b)/(16*(a - b)^2*(1 + Sin[c + d*x])))/d
3.5.10.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[Expa ndIntegrand[(d + e*x^2)^q/(a + c*x^4), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
Time = 3.40 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.29
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -11 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a^{2}-6 a b +35 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a -8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a -11 b}{16 \left (a -b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a^{2}+6 a b -35 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a -b \right )^{3}}+\frac {b^{2} \left (\frac {\left (-3 a -b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}-\frac {\left (-a -3 b \right ) \left (2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a -b \right )^{3}}}{d}\) | \(322\) |
default | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -11 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a^{2}-6 a b +35 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a -8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a -11 b}{16 \left (a -b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a^{2}+6 a b -35 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a -b \right )^{3}}+\frac {b^{2} \left (\frac {\left (-3 a -b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}-\frac {\left (-a -3 b \right ) \left (2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a -b \right )^{3}}}{d}\) | \(322\) |
risch | \(\text {Expression too large to display}\) | \(1171\) |
1/d*(-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(3*a-11*b)/(a-b)^2/(1+sin(d*x+c) )+1/16*(3*a^2-6*a*b+35*b^2)/(a-b)^3*ln(1+sin(d*x+c))+1/2/(8*a-8*b)/(sin(d* x+c)-1)^2-1/16*(3*a-11*b)/(a-b)^2/(sin(d*x+c)-1)+1/16/(a-b)^3*(-3*a^2+6*a* b-35*b^2)*ln(sin(d*x+c)-1)+b^2/(a-b)^3*(1/4*(-3*a-b)*(1/b*a)^(1/4)/a*(ln(( sin(d*x+c)+(1/b*a)^(1/4))/(sin(d*x+c)-(1/b*a)^(1/4)))+2*arctan(sin(d*x+c)/ (1/b*a)^(1/4)))-1/4*(-a-3*b)/b/(1/b*a)^(1/4)*(2*arctan(sin(d*x+c)/(1/b*a)^ (1/4))-ln((sin(d*x+c)+(1/b*a)^(1/4))/(sin(d*x+c)-(1/b*a)^(1/4))))))
Leaf count of result is larger than twice the leaf count of optimal. 3703 vs. \(2 (207) = 414\).
Time = 1.65 (sec) , antiderivative size = 3703, normalized size of antiderivative = 14.87 \[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
-1/16*(4*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*sqrt((6*a^2*b^3 + 20*a*b^4 + 6* b^5 + (a^7 - 6*a^6*b + 15*a^5*b^2 - 20*a^4*b^3 + 15*a^3*b^4 - 6*a^2*b^5 + a*b^6)*d^2*sqrt((a^6*b^5 + 30*a^5*b^6 + 255*a^4*b^7 + 452*a^3*b^8 + 255*a^ 2*b^9 + 30*a*b^10 + b^11)/((a^15 - 12*a^14*b + 66*a^13*b^2 - 220*a^12*b^3 + 495*a^11*b^4 - 792*a^10*b^5 + 924*a^9*b^6 - 792*a^8*b^7 + 495*a^7*b^8 - 220*a^6*b^9 + 66*a^5*b^10 - 12*a^4*b^11 + a^3*b^12)*d^4)))/((a^7 - 6*a^6*b + 15*a^5*b^2 - 20*a^4*b^3 + 15*a^3*b^4 - 6*a^2*b^5 + a*b^6)*d^2))*cos(d*x + c)^4*log(1/2*(a^3*b^4 + 15*a^2*b^5 + 15*a*b^6 + b^7)*sin(d*x + c) + 1/2 *((a^10 - 3*a^9*b - 3*a^8*b^2 + 25*a^7*b^3 - 45*a^6*b^4 + 39*a^5*b^5 - 17* a^4*b^6 + 3*a^3*b^7)*d^3*sqrt((a^6*b^5 + 30*a^5*b^6 + 255*a^4*b^7 + 452*a^ 3*b^8 + 255*a^2*b^9 + 30*a*b^10 + b^11)/((a^15 - 12*a^14*b + 66*a^13*b^2 - 220*a^12*b^3 + 495*a^11*b^4 - 792*a^10*b^5 + 924*a^9*b^6 - 792*a^8*b^7 + 495*a^7*b^8 - 220*a^6*b^9 + 66*a^5*b^10 - 12*a^4*b^11 + a^3*b^12)*d^4)) - (3*a^5*b^3 + 46*a^4*b^4 + 60*a^3*b^5 + 18*a^2*b^6 + a*b^7)*d)*sqrt((6*a^2* b^3 + 20*a*b^4 + 6*b^5 + (a^7 - 6*a^6*b + 15*a^5*b^2 - 20*a^4*b^3 + 15*a^3 *b^4 - 6*a^2*b^5 + a*b^6)*d^2*sqrt((a^6*b^5 + 30*a^5*b^6 + 255*a^4*b^7 + 4 52*a^3*b^8 + 255*a^2*b^9 + 30*a*b^10 + b^11)/((a^15 - 12*a^14*b + 66*a^13* b^2 - 220*a^12*b^3 + 495*a^11*b^4 - 792*a^10*b^5 + 924*a^9*b^6 - 792*a^8*b ^7 + 495*a^7*b^8 - 220*a^6*b^9 + 66*a^5*b^10 - 12*a^4*b^11 + a^3*b^12)*d^4 )))/((a^7 - 6*a^6*b + 15*a^5*b^2 - 20*a^4*b^3 + 15*a^3*b^4 - 6*a^2*b^5 ...
\[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{a - b \sin ^{4}{\left (c + d x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {4 \, b^{2} {\left (\frac {2 \, {\left (b {\left (3 \, \sqrt {a} - \sqrt {b}\right )} + a^{\frac {3}{2}} - 3 \, a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {{\left (b {\left (3 \, \sqrt {a} + \sqrt {b}\right )} + a^{\frac {3}{2}} + 3 \, a \sqrt {b}\right )} \log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} - 6 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (3 \, a^{2} - 6 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {2 \, {\left ({\left (3 \, a - 11 \, b\right )} \sin \left (d x + c\right )^{3} - {\left (5 \, a - 13 \, b\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2} - 2 \, a b + b^{2}}}{16 \, d} \]
1/16*(4*b^2*(2*(b*(3*sqrt(a) - sqrt(b)) + a^(3/2) - 3*a*sqrt(b))*arctan(sq rt(b)*sin(d*x + c)/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*s qrt(b)) + (b*(3*sqrt(a) + sqrt(b)) + a^(3/2) + 3*a*sqrt(b))*log((sqrt(b)*s in(d*x + c) - sqrt(sqrt(a)*sqrt(b)))/(sqrt(b)*sin(d*x + c) + sqrt(sqrt(a)* sqrt(b))))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b)))/(a^3 - 3*a^2*b + 3*a*b ^2 - b^3) + (3*a^2 - 6*a*b + 35*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a^2 - 6*a*b + 35*b^2)*log(sin(d*x + c) - 1)/(a^3 - 3 *a^2*b + 3*a*b^2 - b^3) - 2*((3*a - 11*b)*sin(d*x + c)^3 - (5*a - 13*b)*si n(d*x + c))/((a^2 - 2*a*b + b^2)*sin(d*x + c)^4 - 2*(a^2 - 2*a*b + b^2)*si n(d*x + c)^2 + a^2 - 2*a*b + b^2))/d
Leaf count of result is larger than twice the leaf count of optimal. 630 vs. \(2 (207) = 414\).
Time = 1.28 (sec) , antiderivative size = 630, normalized size of antiderivative = 2.53 \[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\frac {8 \, {\left (\left (-a b^{3}\right )^{\frac {3}{4}} {\left (a + 3 \, b\right )} + \left (-a b^{3}\right )^{\frac {1}{4}} {\left (3 \, a b^{2} + b^{3}\right )}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\sqrt {2} a^{4} b - 3 \, \sqrt {2} a^{3} b^{2} + 3 \, \sqrt {2} a^{2} b^{3} - \sqrt {2} a b^{4}} + \frac {8 \, {\left (\left (-a b^{3}\right )^{\frac {3}{4}} {\left (a + 3 \, b\right )} + \left (-a b^{3}\right )^{\frac {1}{4}} {\left (3 \, a b^{2} + b^{3}\right )}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\sqrt {2} a^{4} b - 3 \, \sqrt {2} a^{3} b^{2} + 3 \, \sqrt {2} a^{2} b^{3} - \sqrt {2} a b^{4}} - \frac {4 \, {\left (\left (-a b^{3}\right )^{\frac {3}{4}} {\left (a + 3 \, b\right )} - \left (-a b^{3}\right )^{\frac {1}{4}} {\left (3 \, a b^{2} + b^{3}\right )}\right )} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{\sqrt {2} a^{4} b - 3 \, \sqrt {2} a^{3} b^{2} + 3 \, \sqrt {2} a^{2} b^{3} - \sqrt {2} a b^{4}} + \frac {4 \, {\left (\left (-a b^{3}\right )^{\frac {3}{4}} {\left (a + 3 \, b\right )} - \left (-a b^{3}\right )^{\frac {1}{4}} {\left (3 \, a b^{2} + b^{3}\right )}\right )} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{\sqrt {2} a^{4} b - 3 \, \sqrt {2} a^{3} b^{2} + 3 \, \sqrt {2} a^{2} b^{3} - \sqrt {2} a b^{4}} - \frac {{\left (3 \, a^{2} - 6 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} - 6 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{3} - 11 \, b \sin \left (d x + c\right )^{3} - 5 \, a \sin \left (d x + c\right ) + 13 \, b \sin \left (d x + c\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*(8*((-a*b^3)^(3/4)*(a + 3*b) + (-a*b^3)^(1/4)*(3*a*b^2 + b^3))*arcta n(1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) + 2*sin(d*x + c))/(-a/b)^(1/4))/(sqrt( 2)*a^4*b - 3*sqrt(2)*a^3*b^2 + 3*sqrt(2)*a^2*b^3 - sqrt(2)*a*b^4) + 8*((-a *b^3)^(3/4)*(a + 3*b) + (-a*b^3)^(1/4)*(3*a*b^2 + b^3))*arctan(-1/2*sqrt(2 )*(sqrt(2)*(-a/b)^(1/4) - 2*sin(d*x + c))/(-a/b)^(1/4))/(sqrt(2)*a^4*b - 3 *sqrt(2)*a^3*b^2 + 3*sqrt(2)*a^2*b^3 - sqrt(2)*a*b^4) - 4*((-a*b^3)^(3/4)* (a + 3*b) - (-a*b^3)^(1/4)*(3*a*b^2 + b^3))*log(sin(d*x + c)^2 + sqrt(2)*( -a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(sqrt(2)*a^4*b - 3*sqrt(2)*a^3*b^2 + 3*sqrt(2)*a^2*b^3 - sqrt(2)*a*b^4) + 4*((-a*b^3)^(3/4)*(a + 3*b) - (-a*b ^3)^(1/4)*(3*a*b^2 + b^3))*log(sin(d*x + c)^2 - sqrt(2)*(-a/b)^(1/4)*sin(d *x + c) + sqrt(-a/b))/(sqrt(2)*a^4*b - 3*sqrt(2)*a^3*b^2 + 3*sqrt(2)*a^2*b ^3 - sqrt(2)*a*b^4) - (3*a^2 - 6*a*b + 35*b^2)*log(abs(sin(d*x + c) + 1))/ (a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 - 6*a*b + 35*b^2)*log(abs(sin(d*x + c) - 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 2*(3*a*sin(d*x + c)^3 - 11*b *sin(d*x + c)^3 - 5*a*sin(d*x + c) + 13*b*sin(d*x + c))/((a^2 - 2*a*b + b^ 2)*(sin(d*x + c)^2 - 1)^2))/d
Time = 20.00 (sec) , antiderivative size = 12217, normalized size of antiderivative = 49.06 \[ \int \frac {\sec ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
(atan(((((18064*a*b^13 + 256*b^14 + 119760*a^2*b^12 - 275888*a^3*b^11 + 11 6624*a^4*b^10 + 28848*a^5*b^9 - 13712*a^6*b^8 + 6768*a^7*b^7 - 720*a^8*b^6 )/(64*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^ 4 - 56*a^5*b^3 + 28*a^6*b^2)) - (((4096*a*b^15 + 12288*a^2*b^14 - 251904*a ^3*b^13 + 1087488*a^4*b^12 - 2457600*a^5*b^11 + 3440640*a^6*b^10 - 3182592 *a^7*b^9 + 2002944*a^8*b^8 - 872448*a^9*b^7 + 266240*a^10*b^6 - 55296*a^11 *b^5 + 6144*a^12*b^4)/(64*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56 *a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)) - (sin(c + d*x)*(-(a^3*( a^3*b^5)^(1/2) + b^3*(a^3*b^5)^(1/2) - 6*a^2*b^5 - 20*a^3*b^4 - 6*a^4*b^3 + 15*a*b^2*(a^3*b^5)^(1/2) + 15*a^2*b*(a^3*b^5)^(1/2))/(16*(a^9 - 6*a^8*b + a^3*b^6 - 6*a^4*b^5 + 15*a^5*b^4 - 20*a^6*b^3 + 15*a^7*b^2)))^(1/2)*(819 2*a^2*b^15 - 73728*a^3*b^14 + 286720*a^4*b^13 - 614400*a^5*b^12 + 737280*a ^6*b^11 - 344064*a^7*b^10 - 344064*a^8*b^9 + 737280*a^9*b^8 - 614400*a^10* b^7 + 286720*a^11*b^6 - 73728*a^12*b^5 + 8192*a^13*b^4))/(16*(a^8 - 8*a^7* b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28 *a^6*b^2)))*(-(a^3*(a^3*b^5)^(1/2) + b^3*(a^3*b^5)^(1/2) - 6*a^2*b^5 - 20* a^3*b^4 - 6*a^4*b^3 + 15*a*b^2*(a^3*b^5)^(1/2) + 15*a^2*b*(a^3*b^5)^(1/2)) /(16*(a^9 - 6*a^8*b + a^3*b^6 - 6*a^4*b^5 + 15*a^5*b^4 - 20*a^6*b^3 + 15*a ^7*b^2)))^(1/2) - (sin(c + d*x)*(256*b^15 - 50464*a^2*b^13 + 190720*a^3*b^ 12 - 280960*a^4*b^11 + 212736*a^5*b^10 - 111296*a^6*b^9 + 57088*a^7*b^8...