Integrand size = 16, antiderivative size = 179 \[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b+b \tan ^2(c+d x)}}\right )}{a^{7/2} d}-\frac {b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}-\frac {b (9 a+5 b) \tan (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 (a+b)^3 d \sqrt {a+b+b \tan ^2(c+d x)}} \]
arctan(a^(1/2)*tan(d*x+c)/(a+b+b*tan(d*x+c)^2)^(1/2))/a^(7/2)/d-1/15*b*(33 *a^2+40*a*b+15*b^2)*tan(d*x+c)/a^3/(a+b)^3/d/(a+b+b*tan(d*x+c)^2)^(1/2)-1/ 5*b*tan(d*x+c)/a/(a+b)/d/(a+b+b*tan(d*x+c)^2)^(5/2)-1/15*b*(9*a+5*b)*tan(d *x+c)/a^2/(a+b)^2/d/(a+b+b*tan(d*x+c)^2)^(3/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 16.81 (sec) , antiderivative size = 1777, normalized size of antiderivative = 9.93 \[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx =\text {Too large to display} \]
(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/ (a + b)]*Cos[c + d*x]^6*Sin[c + d*x])/(8*Sqrt[2]*d*(a + b*Sec[c + d*x]^2)^ (7/2)*(a + b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3 /2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a*AppellF1[3/2, -3, 9 /2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[ 3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d* x]^2)*((21*a*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^7*Sin[c + d*x]^2)/(8*Sqrt[2]*(a + b - a*Si n[c + d*x]^2)^(9/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a*AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d* x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d*x]^2)) + (3*(a + b) *AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*C os[c + d*x]^7)/(8*Sqrt[2]*(a + b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*Appe llF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a *AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/ (a + b)])*Sin[c + d*x]^2)) - (9*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^5*Sin[c + d*x]^2)/(4*Sq rt[2]*(a + b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2...
Time = 0.33 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4616, 316, 402, 402, 27, 291, 216}
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.
| \(\displaystyle \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \sec (c+d x)^2\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4616 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a+b\right )^{7/2}}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-4 b \tan ^2(c+d x)+5 a+b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a+b\right )^{5/2}}d\tan (c+d x)}{5 a (a+b)}-\frac {b \tan (c+d x)}{5 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^2+12 b a+5 b^2-2 b (9 a+5 b) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a+b\right )^{3/2}}d\tan (c+d x)}{3 a (a+b)}-\frac {b (9 a+5 b) \tan (c+d x)}{3 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tan (c+d x)}{5 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {15 (a+b)^3}{\left (\tan ^2(c+d x)+1\right ) \sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)}{a (a+b)}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{a (a+b) \sqrt {a+b \tan ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tan (c+d x)}{3 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tan (c+d x)}{5 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{a (a+b) \sqrt {a+b \tan ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tan (c+d x)}{3 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tan (c+d x)}{5 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\frac {a \tan ^2(c+d x)}{b \tan ^2(c+d x)+a+b}+1}d\frac {\tan (c+d x)}{\sqrt {b \tan ^2(c+d x)+a+b}}}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{a (a+b) \sqrt {a+b \tan ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tan (c+d x)}{3 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tan (c+d x)}{5 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {15 (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )}{a^{3/2}}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{a (a+b) \sqrt {a+b \tan ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tan (c+d x)}{3 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tan (c+d x)}{5 a (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}}}{d}\) |
(-1/5*(b*Tan[c + d*x])/(a*(a + b)*(a + b + b*Tan[c + d*x]^2)^(5/2)) + (-1/ 3*(b*(9*a + 5*b)*Tan[c + d*x])/(a*(a + b)*(a + b + b*Tan[c + d*x]^2)^(3/2) ) + ((15*(a + b)^2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + b + b*Tan[c + d* x]^2]])/a^(3/2) - (b*(33*a^2 + 40*a*b + 15*b^2)*Tan[c + d*x])/(a*(a + b)*S qrt[a + b + b*Tan[c + d*x]^2]))/(3*a*(a + b)))/(5*a*(a + b)))/d
3.3.96.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(2017\) vs. \(2(161)=322\).
Time = 5.46 (sec) , antiderivative size = 2018, normalized size of antiderivative = 11.27
-1/15/d/(a+b)^3/a^3/(-a)^(1/2)*(a*(1-cos(d*x+c))^4*csc(d*x+c)^4+b*(1-cos(d *x+c))^4*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c)) ^2*csc(d*x+c)^2+a+b)*(-30*(-a)^(1/2)*b^5*(-cot(d*x+c)+csc(d*x+c))+15*ln(4* ((-a)^(1/2)*(a*(1-cos(d*x+c))^4*csc(d*x+c)^4+b*(1-cos(d*x+c))^4*csc(d*x+c) ^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b )^(1/2)-2*a*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))*( a*(1-cos(d*x+c))^4*csc(d*x+c)^4+b*(1-cos(d*x+c))^4*csc(d*x+c)^4-2*a*(1-cos (d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(5/2)*a^3+1 5*ln(4*((-a)^(1/2)*(a*(1-cos(d*x+c))^4*csc(d*x+c)^4+b*(1-cos(d*x+c))^4*csc (d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c )^2+a+b)^(1/2)-2*a*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^ 2+1))*(a*(1-cos(d*x+c))^4*csc(d*x+c)^4+b*(1-cos(d*x+c))^4*csc(d*x+c)^4-2*a *(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(5/2 )*b^3-80*(-a)^(1/2)*a^2*b^3*(1-cos(d*x+c))^7*csc(d*x+c)^7-320*(-a)^(1/2)*a *b^4*(1-cos(d*x+c))^7*csc(d*x+c)^7-340*(-a)^(1/2)*a*b^4*(1-cos(d*x+c))^5*c sc(d*x+c)^5+480*(-a)^(1/2)*a^3*b^2*(1-cos(d*x+c))^3*csc(d*x+c)^3-80*(-a)^( 1/2)*a^2*b^3*(1-cos(d*x+c))^3*csc(d*x+c)^3+360*(-a)^(1/2)*a^4*b*(1-cos(d*x +c))^3*csc(d*x+c)^3-320*(-a)^(1/2)*a*b^4*(1-cos(d*x+c))^3*csc(d*x+c)^3-300 *(-a)^(1/2)*a^2*b^3*(1-cos(d*x+c))^9*csc(d*x+c)^9-150*(-a)^(1/2)*a*b^4*(1- cos(d*x+c))^9*csc(d*x+c)^9+480*(-a)^(1/2)*a^3*b^2*(1-cos(d*x+c))^7*csc(...
Leaf count of result is larger than twice the leaf count of optimal. 560 vs. \(2 (161) = 322\).
Time = 2.12 (sec) , antiderivative size = 1241, normalized size of antiderivative = 6.93 \[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]
[-1/120*(15*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(d*x + c)^6 + a^3*b^ 3 + 3*a^2*b^4 + 3*a*b^5 + b^6 + 3*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4 )*cos(d*x + c)^4 + 3*(a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*cos(d*x + c )^2)*sqrt(-a)*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 - a^3*b)*cos(d*x + c)^ 6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(d*x + c)^4 + a^4 - 28*a^3*b + 70 *a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(d*x + c)^2 + 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 - a^2*b)*cos(d*x + c)^5 + 2*( 5*a^3 - 14*a^2*b + 5*a*b^2)*cos(d*x + c)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^ 3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)*sin( d*x + c)) + 8*((45*a^5*b + 60*a^4*b^2 + 23*a^3*b^3)*cos(d*x + c)^5 + (75*a ^4*b^2 + 94*a^3*b^3 + 35*a^2*b^4)*cos(d*x + c)^3 + (33*a^3*b^3 + 40*a^2*b^ 4 + 15*a*b^5)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)*si n(d*x + c))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7*b^3)*d*cos(d*x + c)^6 + 3*( a^9*b + 3*a^8*b^2 + 3*a^7*b^3 + a^6*b^4)*d*cos(d*x + c)^4 + 3*(a^8*b^2 + 3 *a^7*b^3 + 3*a^6*b^4 + a^5*b^5)*d*cos(d*x + c)^2 + (a^7*b^3 + 3*a^6*b^4 + 3*a^5*b^5 + a^4*b^6)*d), -1/60*(15*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)* cos(d*x + c)^6 + a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6 + 3*(a^5*b + 3*a^4*b^ 2 + 3*a^3*b^3 + a^2*b^4)*cos(d*x + c)^4 + 3*(a^4*b^2 + 3*a^3*b^3 + 3*a^2*b ^4 + a*b^5)*cos(d*x + c)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(d*x + c)^5 - 8*( a^2 - a*b)*cos(d*x + c)^3 + (a^2 - 6*a*b + b^2)*cos(d*x + c))*sqrt(a)*s...
\[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \sec ^{2}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
Timed out. \[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right )^{2} + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\cos \left (c+d\,x\right )}^2}\right )}^{7/2}} \,d x \]