Integrand size = 23, antiderivative size = 747 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=-\frac {b^{5/3} \left (4 a^2-3 a^{4/3} b^{2/3}+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \left (a^2-b^2\right )^2 d}-\frac {b^{5/3} \left (4 a^{8/3}-9 a^2 b^{2/3}+8 a^{2/3} b^2-3 b^{8/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} \left (a^2-b^2\right )^3 d}-\frac {(a+7 b) \log (1-\sin (c+d x))}{4 (a+b)^3 d}+\frac {(a-7 b) \log (1+\sin (c+d x))}{4 (a-b)^3 d}+\frac {b^{5/3} \left (4 a^2+3 a^{4/3} b^{2/3}+2 b^2\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \left (a^2-b^2\right )^2 d}+\frac {b^{5/3} \left (3 b^{2/3} \left (3 a^2+b^2\right )+4 a^{2/3} \left (a^2+2 b^2\right )\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 \sqrt [3]{a} \left (a^2-b^2\right )^3 d}-\frac {b^{5/3} \left (4 a^2+3 a^{4/3} b^{2/3}+2 b^2\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{18 a^{5/3} \left (a^2-b^2\right )^2 d}-\frac {b^{5/3} \left (3 b^{2/3} \left (3 a^2+b^2\right )+4 a^{2/3} \left (a^2+2 b^2\right )\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 \sqrt [3]{a} \left (a^2-b^2\right )^3 d}+\frac {2 a b \left (a^2+5 b^2\right ) \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac {1}{4 (a+b)^2 d (1-\sin (c+d x))}-\frac {1}{4 (a-b)^2 d (1+\sin (c+d x))}-\frac {b \left (a \left (a^2+2 b^2\right )-b \sin (c+d x) \left (2 a^2+b^2-3 a b \sin (c+d x)\right )\right )}{3 a \left (a^2-b^2\right )^2 d \left (a+b \sin ^3(c+d x)\right )} \]
-1/4*(a+7*b)*ln(1-sin(d*x+c))/(a+b)^3/d+1/4*(a-7*b)*ln(1+sin(d*x+c))/(a-b) ^3/d+1/9*b^(5/3)*(4*a^2+3*a^(4/3)*b^(2/3)+2*b^2)*ln(a^(1/3)+b^(1/3)*sin(d* x+c))/a^(5/3)/(a^2-b^2)^2/d+1/3*b^(5/3)*(3*b^(2/3)*(3*a^2+b^2)+4*a^(2/3)*( a^2+2*b^2))*ln(a^(1/3)+b^(1/3)*sin(d*x+c))/a^(1/3)/(a^2-b^2)^3/d-1/18*b^(5 /3)*(4*a^2+3*a^(4/3)*b^(2/3)+2*b^2)*ln(a^(2/3)-a^(1/3)*b^(1/3)*sin(d*x+c)+ b^(2/3)*sin(d*x+c)^2)/a^(5/3)/(a^2-b^2)^2/d-1/6*b^(5/3)*(3*b^(2/3)*(3*a^2+ b^2)+4*a^(2/3)*(a^2+2*b^2))*ln(a^(2/3)-a^(1/3)*b^(1/3)*sin(d*x+c)+b^(2/3)* sin(d*x+c)^2)/a^(1/3)/(a^2-b^2)^3/d+2/3*a*b*(a^2+5*b^2)*ln(a+b*sin(d*x+c)^ 3)/(a^2-b^2)^3/d+1/4/(a+b)^2/d/(1-sin(d*x+c))-1/4/(a-b)^2/d/(1+sin(d*x+c)) -1/3*b*(a*(a^2+2*b^2)-b*sin(d*x+c)*(2*a^2+b^2-3*a*b*sin(d*x+c)))/a/(a^2-b^ 2)^2/d/(a+b*sin(d*x+c)^3)-1/9*b^(5/3)*(4*a^2-3*a^(4/3)*b^(2/3)+2*b^2)*arct an(1/3*(a^(1/3)-2*b^(1/3)*sin(d*x+c))/a^(1/3)*3^(1/2))/a^(5/3)/(a^2-b^2)^2 /d*3^(1/2)-1/3*b^(5/3)*(4*a^(8/3)-9*a^2*b^(2/3)+8*a^(2/3)*b^2-3*b^(8/3))*a rctan(1/3*(a^(1/3)-2*b^(1/3)*sin(d*x+c))/a^(1/3)*3^(1/2))/a^(1/3)/(a^2-b^2 )^3/d*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.43 (sec) , antiderivative size = 724, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\frac {-\frac {2 b^{5/3} \left (2 a^2+b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \left (a^2-b^2\right )^2}-\frac {4 \sqrt [3]{a} b^{5/3} \left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \left (a^2-b^2\right )^3}-\frac {(a+7 b) \log (1-\sin (c+d x))}{4 (a+b)^3}+\frac {(a-7 b) \log (1+\sin (c+d x))}{4 (a-b)^3}+\frac {2 b^{5/3} \left (2 a^2+b^2\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \left (a^2-b^2\right )^2}+\frac {4 \sqrt [3]{a} b^{5/3} \left (a^2+2 b^2\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3}-\frac {b^{5/3} \left (2 a^2+b^2\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} \left (a^2-b^2\right )^2}-\frac {2 \sqrt [3]{a} b^{5/3} \left (a^2+2 b^2\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{3 \left (a^2-b^2\right )^3}+\frac {2 a b \left (a^2+5 b^2\right ) \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right )^3}+\frac {1}{4 (a+b)^2 (1-\sin (c+d x))}-\frac {3 b^3 \left (3 a^2+b^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b \sin ^3(c+d x)}{a}\right ) \sin ^2(c+d x)}{2 a \left (a^2-b^2\right )^3}-\frac {3 b^3 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},2,\frac {5}{3},-\frac {b \sin ^3(c+d x)}{a}\right ) \sin ^2(c+d x)}{2 a \left (a^2-b^2\right )^2}-\frac {1}{4 (a-b)^2 (1+\sin (c+d x))}-\frac {b \left (a^2+2 b^2\right )}{3 \left (a^2-b^2\right )^2 \left (a+b \sin ^3(c+d x)\right )}+\frac {a b^2 \left (2+\frac {b^2}{a^2}\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 \left (a+b \sin ^3(c+d x)\right )}}{d} \]
((-2*b^(5/3)*(2*a^2 + b^2)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt [3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*(a^2 - b^2)^2) - (4*a^(1/3)*b^(5/3)*(a^2 + 2*b^2)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(S qrt[3]*(a^2 - b^2)^3) - ((a + 7*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^3) + ((a - 7*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^3) + (2*b^(5/3)*(2*a^2 + b^2) *Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*(a^2 - b^2)^2) + (4*a^(1/ 3)*b^(5/3)*(a^2 + 2*b^2)*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(3*(a^2 - b^ 2)^3) - (b^(5/3)*(2*a^2 + b^2)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2])/(9*a^(5/3)*(a^2 - b^2)^2) - (2*a^(1/3)*b^(5/3)* (a^2 + 2*b^2)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2])/(3*(a^2 - b^2)^3) + (2*a*b*(a^2 + 5*b^2)*Log[a + b*Sin[c + d*x]^ 3])/(3*(a^2 - b^2)^3) + 1/(4*(a + b)^2*(1 - Sin[c + d*x])) - (3*b^3*(3*a^2 + b^2)*Hypergeometric2F1[2/3, 1, 5/3, -((b*Sin[c + d*x]^3)/a)]*Sin[c + d* x]^2)/(2*a*(a^2 - b^2)^3) - (3*b^3*Hypergeometric2F1[2/3, 2, 5/3, -((b*Sin [c + d*x]^3)/a)]*Sin[c + d*x]^2)/(2*a*(a^2 - b^2)^2) - 1/(4*(a - b)^2*(1 + Sin[c + d*x])) - (b*(a^2 + 2*b^2))/(3*(a^2 - b^2)^2*(a + b*Sin[c + d*x]^3 )) + (a*b^2*(2 + b^2/a^2)*Sin[c + d*x])/(3*(a^2 - b^2)^2*(a + b*Sin[c + d* x]^3)))/d
Time = 1.20 (sec) , antiderivative size = 715, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3702, 7293, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 \left (a+b \sin (c+d x)^3\right )^2}dx\) |
| \(\Big \downarrow \) 3702 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\sin ^2(c+d x)\right )^2 \left (b \sin ^3(c+d x)+a\right )^2}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {\int \left (\frac {\left (2 a \left (a^2+5 b^2\right ) \sin ^2(c+d x)-3 b \left (3 a^2+b^2\right ) \sin (c+d x)+4 a \left (a^2+2 b^2\right )\right ) b^2}{\left (a^2-b^2\right )^3 \left (b \sin ^3(c+d x)+a\right )}+\frac {\left (2 a^2-3 b \sin (c+d x) a+b^2+\left (a^2+2 b^2\right ) \sin ^2(c+d x)\right ) b^2}{\left (a^2-b^2\right )^2 \left (b \sin ^3(c+d x)+a\right )^2}+\frac {-a-7 b}{4 (a+b)^3 (\sin (c+d x)-1)}+\frac {a-7 b}{4 (a-b)^3 (\sin (c+d x)+1)}+\frac {1}{4 (a+b)^2 (\sin (c+d x)-1)^2}+\frac {1}{4 (a-b)^2 (\sin (c+d x)+1)^2}\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b \left (a \left (a^2+2 b^2\right )-b \sin (c+d x) \left (2 a^2-3 a b \sin (c+d x)+b^2\right )\right )}{3 a \left (a^2-b^2\right )^2 \left (a+b \sin ^3(c+d x)\right )}+\frac {2 a b \left (a^2+5 b^2\right ) \log \left (a+b \sin ^3(c+d x)\right )}{3 \left (a^2-b^2\right )^3}-\frac {b^{5/3} \left (-3 a^{4/3} b^{2/3}+4 a^2+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \left (a^2-b^2\right )^2}-\frac {b^{5/3} \left (8 a^{2/3} b^2+4 a^{8/3}-9 a^2 b^{2/3}-3 b^{8/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} \left (a^2-b^2\right )^3}-\frac {b^{5/3} \left (3 a^{4/3} b^{2/3}+4 a^2+2 b^2\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{18 a^{5/3} \left (a^2-b^2\right )^2}-\frac {b^{5/3} \left (3 b^{2/3} \left (3 a^2+b^2\right )+4 a^{2/3} \left (a^2+2 b^2\right )\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 \sqrt [3]{a} \left (a^2-b^2\right )^3}+\frac {b^{5/3} \left (3 a^{4/3} b^{2/3}+4 a^2+2 b^2\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \left (a^2-b^2\right )^2}+\frac {b^{5/3} \left (3 b^{2/3} \left (3 a^2+b^2\right )+4 a^{2/3} \left (a^2+2 b^2\right )\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 \sqrt [3]{a} \left (a^2-b^2\right )^3}+\frac {1}{4 (a+b)^2 (1-\sin (c+d x))}-\frac {1}{4 (a-b)^2 (\sin (c+d x)+1)}-\frac {(a+7 b) \log (1-\sin (c+d x))}{4 (a+b)^3}+\frac {(a-7 b) \log (\sin (c+d x)+1)}{4 (a-b)^3}}{d}\) |
(-1/3*(b^(5/3)*(4*a^2 - 3*a^(4/3)*b^(2/3) + 2*b^2)*ArcTan[(a^(1/3) - 2*b^( 1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3)*(a^2 - b^2)^2) - ( b^(5/3)*(4*a^(8/3) - 9*a^2*b^(2/3) + 8*a^(2/3)*b^2 - 3*b^(8/3))*ArcTan[(a^ (1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(1/3)*(a^2 - b^2)^3) - ((a + 7*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^3) + ((a - 7*b)*L og[1 + Sin[c + d*x]])/(4*(a - b)^3) + (b^(5/3)*(4*a^2 + 3*a^(4/3)*b^(2/3) + 2*b^2)*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*(a^2 - b^2)^2) + (b^(5/3)*(3*b^(2/3)*(3*a^2 + b^2) + 4*a^(2/3)*(a^2 + 2*b^2))*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(3*a^(1/3)*(a^2 - b^2)^3) - (b^(5/3)*(4*a^2 + 3*a^ (4/3)*b^(2/3) + 2*b^2)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3 )*Sin[c + d*x]^2])/(18*a^(5/3)*(a^2 - b^2)^2) - (b^(5/3)*(3*b^(2/3)*(3*a^2 + b^2) + 4*a^(2/3)*(a^2 + 2*b^2))*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d *x] + b^(2/3)*Sin[c + d*x]^2])/(6*a^(1/3)*(a^2 - b^2)^3) + (2*a*b*(a^2 + 5 *b^2)*Log[a + b*Sin[c + d*x]^3])/(3*(a^2 - b^2)^3) + 1/(4*(a + b)^2*(1 - S in[c + d*x])) - 1/(4*(a - b)^2*(1 + Sin[c + d*x])) - (b*(a*(a^2 + 2*b^2) - b*Sin[c + d*x]*(2*a^2 + b^2 - 3*a*b*Sin[c + d*x])))/(3*a*(a^2 - b^2)^2*(a + b*Sin[c + d*x]^3)))/d
3.4.98.3.1 Defintions of rubi rules used
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 = 4.78 (sec) , antiderivative size = 483, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a -7 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {1}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -7 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}+\frac {b^{2} \left (\frac {\left (-a^{2} b +b^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )+\frac {\left (2 a^{4}-a^{2} b^{2}-b^{4}\right ) \sin \left (d x +c \right )}{3 a}-\frac {a^{4}+a^{2} b^{2}-2 b^{4}}{3 b}}{a +b \left (\sin ^{3}\left (d x +c \right )\right )}+\frac {\frac {2 \left (8 a^{4}+11 a^{2} b^{2}-b^{4}\right ) \left (\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {2 \left (-15 a^{3} b -3 a \,b^{3}\right ) \left (-\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}+\frac {2 \left (3 a^{4}+15 a^{2} b^{2}\right ) \ln \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}{9 b}}{a}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\) | \(483\) |
| default | \(\frac {-\frac {1}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a -7 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {1}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -7 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}+\frac {b^{2} \left (\frac {\left (-a^{2} b +b^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )+\frac {\left (2 a^{4}-a^{2} b^{2}-b^{4}\right ) \sin \left (d x +c \right )}{3 a}-\frac {a^{4}+a^{2} b^{2}-2 b^{4}}{3 b}}{a +b \left (\sin ^{3}\left (d x +c \right )\right )}+\frac {\frac {2 \left (8 a^{4}+11 a^{2} b^{2}-b^{4}\right ) \left (\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {2 \left (-15 a^{3} b -3 a \,b^{3}\right ) \left (-\frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}+\frac {2 \left (3 a^{4}+15 a^{2} b^{2}\right ) \ln \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}{9 b}}{a}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\) | \(483\) |
| risch | \(\text {Expression too large to display}\) | \(1831\) |
1/d*(-1/4/(a+b)^2/(sin(d*x+c)-1)+1/4/(a+b)^3*(-a-7*b)*ln(sin(d*x+c)-1)-1/4 /(a-b)^2/(1+sin(d*x+c))+1/4*(a-7*b)/(a-b)^3*ln(1+sin(d*x+c))+b^2/(a-b)^3/( a+b)^3*(((-a^2*b+b^3)*sin(d*x+c)^2+1/3*(2*a^4-a^2*b^2-b^4)/a*sin(d*x+c)-1/ 3*(a^4+a^2*b^2-2*b^4)/b)/(a+b*sin(d*x+c)^3)+2/3/a*((8*a^4+11*a^2*b^2-b^4)* (1/3/b/(1/b*a)^(2/3)*ln(sin(d*x+c)+(1/b*a)^(1/3))-1/6/b/(1/b*a)^(2/3)*ln(s in(d*x+c)^2-(1/b*a)^(1/3)*sin(d*x+c)+(1/b*a)^(2/3))+1/3/b/(1/b*a)^(2/3)*3^ (1/2)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*sin(d*x+c)-1)))+(-15*a^3*b-3*a*b ^3)*(-1/3/b/(1/b*a)^(1/3)*ln(sin(d*x+c)+(1/b*a)^(1/3))+1/6/b/(1/b*a)^(1/3) *ln(sin(d*x+c)^2-(1/b*a)^(1/3)*sin(d*x+c)+(1/b*a)^(2/3))+1/3*3^(1/2)/b/(1/ b*a)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*sin(d*x+c)-1)))+1/3*(3*a^4+ 15*a^2*b^2)/b*ln(a+b*sin(d*x+c)^3))))
Result contains complex when optimal does not.
Time = 7.04 (sec) , antiderivative size = 15989, normalized size of antiderivative = 21.40 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 788, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Too large to display} \]
-1/36*(8*sqrt(3)*(5*a^3*b^2*(3*(a/b)^(2/3) + 2) - a^2*b^3*(11*(a/b)^(1/3) + 10*a/b) - 2*a^4*b*(4*(a/b)^(1/3) + a/b) + 3*a*b^4*(a/b)^(2/3) + b^5*(a/b )^(1/3) + 2*a^5)*arctan(-1/3*sqrt(3)*((a/b)^(1/3) - 2*sin(d*x + c))/(a/b)^ (1/3))/((a^7*(a/b)^(2/3) - 3*a^5*b^2*(a/b)^(2/3) + 3*a^3*b^4*(a/b)^(2/3) - a*b^6*(a/b)^(2/3))*(a/b)^(1/3)) - 4*(a^2*b^3*(30*(a/b)^(2/3) - 11) + 2*a^ 4*b*(3*(a/b)^(2/3) - 4) - 15*a^3*b^2*(a/b)^(1/3) - 3*a*b^4*(a/b)^(1/3) + b ^5)*log(sin(d*x + c)^2 - (a/b)^(1/3)*sin(d*x + c) + (a/b)^(2/3))/(a^7*(a/b )^(2/3) - 3*a^5*b^2*(a/b)^(2/3) + 3*a^3*b^4*(a/b)^(2/3) - a*b^6*(a/b)^(2/3 )) - 8*(a^2*b^3*(15*(a/b)^(2/3) + 11) + a^4*b*(3*(a/b)^(2/3) + 8) + 15*a^3 *b^2*(a/b)^(1/3) + 3*a*b^4*(a/b)^(1/3) - b^5)*log((a/b)^(1/3) + sin(d*x + c))/(a^7*(a/b)^(2/3) - 3*a^5*b^2*(a/b)^(2/3) + 3*a^3*b^4*(a/b)^(2/3) - a*b ^6*(a/b)^(2/3)) - 9*(a - 7*b)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b ^2 - b^3) + 9*(a + 7*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b ^3) - 6*(3*(a^3*b + 3*a*b^3)*sin(d*x + c)^4 - 8*a^3*b - 4*a*b^3 - 2*(5*a^2 *b^2 + b^4)*sin(d*x + c)^3 + 2*(a^3*b - a*b^3)*sin(d*x + c)^2 + (3*a^4 + 7 *a^2*b^2 + 2*b^4)*sin(d*x + c))/(a^6 - 2*a^4*b^2 + a^2*b^4 - (a^5*b - 2*a^ 3*b^3 + a*b^5)*sin(d*x + c)^5 + (a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c)^3 - (a^6 - 2*a^4*b^2 + a^2*b^4)*sin(d*x + c)^2))/d
\[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sin \left (d x + c\right )^{3} + a\right )}^{2}} \,d x } \]
Time = 14.27 (sec) , antiderivative size = 1605, normalized size of antiderivative = 2.15 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Too large to display} \]
symsum(log(root(2187*a^9*b^2*z^3 - 2187*a^7*b^4*z^3 + 729*a^5*b^6*z^3 - 72 9*a^11*z^3 + 7290*a^6*b^3*z^2 + 1458*a^8*b*z^2 - 972*a^5*b^2*z + 324*a^3*b ^4*z + 216*a^2*b^3 - 8*b^5, z, k)*(((32*a*b^11)/27 + (2173*a^3*b^9)/27 + ( 847*a^5*b^7)/3 - 20*a^7*b^5)/(a^11 + a^3*b^8 - 4*a^5*b^6 + 6*a^7*b^4 - 4*a ^9*b^2) - root(2187*a^9*b^2*z^3 - 2187*a^7*b^4*z^3 + 729*a^5*b^6*z^3 - 729 *a^11*z^3 + 7290*a^6*b^3*z^2 + 1458*a^8*b*z^2 - 972*a^5*b^2*z + 324*a^3*b^ 4*z + 216*a^2*b^3 - 8*b^5, z, k)*(((32*a^2*b^12)/3 - (1017*a^4*b^10)/4 + 3 25*a^6*b^8 + (4153*a^8*b^6)/12 - (63*a^10*b^4)/2)/(a^11 + a^3*b^8 - 4*a^5* b^6 + 6*a^7*b^4 - 4*a^9*b^2) + root(2187*a^9*b^2*z^3 - 2187*a^7*b^4*z^3 + 729*a^5*b^6*z^3 - 729*a^11*z^3 + 7290*a^6*b^3*z^2 + 1458*a^8*b*z^2 - 972*a ^5*b^2*z + 324*a^3*b^4*z + 216*a^2*b^3 - 8*b^5, z, k)*((16*a^3*b^13 - (563 *a^5*b^11)/2 + 303*a^7*b^9 + 188*a^9*b^7 - 239*a^11*b^5 + (27*a^13*b^3)/2) /(a^11 + a^3*b^8 - 4*a^5*b^6 + 6*a^7*b^4 - 4*a^9*b^2) + root(2187*a^9*b^2* z^3 - 2187*a^7*b^4*z^3 + 729*a^5*b^6*z^3 - 729*a^11*z^3 + 7290*a^6*b^3*z^2 + 1458*a^8*b*z^2 - 972*a^5*b^2*z + 324*a^3*b^4*z + 216*a^2*b^3 - 8*b^5, z , k)*((36*a^4*b^14 + 36*a^6*b^12 - 504*a^8*b^10 + 936*a^10*b^8 - 684*a^12* b^6 + 180*a^14*b^4)/(a^11 + a^3*b^8 - 4*a^5*b^6 + 6*a^7*b^4 - 4*a^9*b^2) + (sin(c + d*x)*(17496*a^5*b^13 - 64152*a^7*b^11 + 81648*a^9*b^9 - 34992*a^ 11*b^7 - 5832*a^13*b^5 + 5832*a^15*b^3))/(108*(a^11 + a^3*b^8 - 4*a^5*b^6 + 6*a^7*b^4 - 4*a^9*b^2))) - (sin(c + d*x)*(13824*a^4*b^12 - 864*a^2*b^...