Integrand size = 25, antiderivative size = 187 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {3 a (a+4 b) \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{7/2} f}-\frac {5 a \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {b (13 a+2 b) \tan (e+f x)}{8 (a-b)^3 f \sqrt {a+b \tan ^2(e+f x)}} \]
3/8*a*(a+4*b)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b )^(7/2)/f-5/8*a*cos(f*x+e)*sin(f*x+e)/(a-b)^2/f/(a+b*tan(f*x+e)^2)^(1/2)+1 /4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)-1/8*b*(13*a+2* b)*tan(f*x+e)/(a-b)^3/f/(a+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 4.59 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.74 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\left (-\left ((a-b) \left (7 a^2+48 a b+5 b^2+\left (6 a^2-2 a b-4 b^2\right ) \cos (2 (e+f x))-(a-b)^2 \cos (4 (e+f x))\right )\right )+6 \sqrt {2} a \left (a^2+3 a b-4 b^2\right ) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )-6 \sqrt {2} a^2 (a+4 b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \sec ^2(e+f x) \sin (2 (e+f x))}{32 \sqrt {2} (a-b)^4 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
((-((a - b)*(7*a^2 + 48*a*b + 5*b^2 + (6*a^2 - 2*a*b - 4*b^2)*Cos[2*(e + f *x)] - (a - b)^2*Cos[4*(e + f*x)])) + 6*Sqrt[2]*a*(a^2 + 3*a*b - 4*b^2)*Sq rt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin [Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - 6*Sqrt[2]*a^2*(a + 4*b)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*( e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x) ])/(32*Sqrt[2]*(a - b)^4*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
Time = 0.38 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4146, 372, 27, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{\left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\int \frac {a \left (1-4 \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {a \int \frac {1-4 \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {a \left (\frac {5 \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\int \frac {-10 b \tan ^2(e+f x)+3 a+2 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{2 (a-b)}\right )}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {a \left (\frac {5 \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\frac {\int \frac {3 a (a+4 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a (a-b)}-\frac {b (13 a+2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}\right )}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {a \left (\frac {5 \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\frac {3 (a+4 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a-b}-\frac {b (13 a+2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}\right )}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {a \left (\frac {5 \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\frac {3 (a+4 b) \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{a-b}-\frac {b (13 a+2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}\right )}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {a \left (\frac {5 \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\frac {3 (a+4 b) \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2}}-\frac {b (13 a+2 b) \tan (e+f x)}{a (a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}\right )}{4 (a-b)}}{f}\) |
(Tan[e + f*x]/(4*(a - b)*(1 + Tan[e + f*x]^2)^2*Sqrt[a + b*Tan[e + f*x]^2] ) - (a*((5*Tan[e + f*x])/(2*(a - b)*(1 + Tan[e + f*x]^2)*Sqrt[a + b*Tan[e + f*x]^2]) - ((3*(a + 4*b)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Ta n[e + f*x]^2]])/(a - b)^(3/2) - (b*(13*a + 2*b)*Tan[e + f*x])/(a*(a - b)*S qrt[a + b*Tan[e + f*x]^2]))/(2*(a - b))))/(4*(a - b)))/f
3.2.34.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 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[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(4725\) vs. \(2(167)=334\).
Time = 4.43 (sec) , antiderivative size = 4726, normalized size of antiderivative = 25.27
1/f/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b* tan(f*x+e)^2)^(1/2)*tan(f*x+e))-b*tan(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^ (1/2)+1/2/f/b/(a-b)^3*(b^4*(a-b))^(1/2)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2) /(b*(-cos(2*f*x+2*e)+1)^2*csc(2*f*x+2*e)^2+a)^(1/2)*(csc(2*f*x+2*e)-cot(2* f*x+2*e)))-1/4/f/(a-b)^2*a/(b*(-cos(2*f*x+2*e)+1)^2*csc(2*f*x+2*e)^2+a)^(1 /2)/(1/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e )^2*b+b*csc(2*f*x+2*e)^2+a)*a*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)^2-2/(cos(2*f *x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f *x+2*e)^2+a)*a*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)+1/(cos(2*f*x+2*e)^2*csc(2*f *x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*a*cs c(2*f*x+2*e)^2-b/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc (2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)^2+2* b/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b +b*csc(2*f*x+2*e)^2+a)*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)-b/(cos(2*f*x+2*e)^2 *csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2 +a)*csc(2*f*x+2*e)^2+1)*csc(2*f*x+2*e)+1/4/f/(a-b)^2*a/(b*(-cos(2*f*x+2*e) +1)^2*csc(2*f*x+2*e)^2+a)^(1/2)/(1/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2* cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*a*csc(2*f*x+2*e)^2 *cos(2*f*x+2*e)^2-2/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)* csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*a*csc(2*f*x+2*e)^2*cos(2*f*x+2...
Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (167) = 334\).
Time = 80.81 (sec) , antiderivative size = 1046, normalized size of antiderivative = 5.59 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/64*(3*(a^2*b + 4*a*b^2 + (a^3 + 3*a^2*b - 4*a*b^2)*cos(f*x + e)^2)*sqrt (-a + b)*log(128*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^ 8 - 256*(a^4 - 5*a^3*b + 9*a^2*b^2 - 7*a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32* (5*a^4 - 34*a^3*b + 77*a^2*b^2 - 72*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 - 32*a^3*b + 160*a^2*b^2 - 256*a*b^3 + 128*b^4 - 32*(a^4 - 11*a^3*b + 34*a^ 2*b^2 - 40*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 24*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cos(f*x + e) ^5 + 2*(5*a^3 - 29*a^2*b + 48*a*b^2 - 24*b^3)*cos(f*x + e)^3 - (a^3 - 10*a ^2*b + 24*a*b^2 - 16*b^3)*cos(f*x + e))*sqrt(-a + b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - 5*(a^3 - 2*a^2*b + a*b^2)*cos(f*x + e)^3 - (13*a^ 2*b - 11*a*b^2 - 2*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/co s(f*x + e)^2)*sin(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5* a*b^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f), 1/32*(3*(a^2*b + 4*a*b^2 + (a^3 + 3*a^2*b - 4*a*b^2)*cos(f*x + e )^2)*sqrt(a - b)*arctan(-1/4*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 8*(a^ 2 - 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 - 8*a*b + 8*b^2)*cos(f*x + e))*sq rt(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*(a^3 - 3*a ^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^4 - a^2*b + 3*a*b^2 - 2*b^3 - (a^3 - 6* a^2*b + 9*a*b^2 - 4*b^3)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(2*(a^3 - 3...
\[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sin ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]