Integrand size = 25, antiderivative size = 172 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {b} (3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a+5 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 (a+b) f}-\frac {(3 a+5 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f} \]
1/2*(3*a+5*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/ 2)/f+1/2*b*(3*a+5*b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/(a+b)/f-1/3*(3* a+5*b)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)/f-1/3*cot(f*x+e)^3*(a+b +b*tan(f*x+e)^2)^(5/2)/(a+b)/f
Result contains complex when optimal does not.
Time = 7.69 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.15 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {2} e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (-\frac {i \left (4 a \left (1+e^{2 i (e+f x)}\right )^2 \left (1-4 e^{2 i (e+f x)}+e^{4 i (e+f x)}\right )+b \left (15-20 e^{2 i (e+f x)}-22 e^{4 i (e+f x)}-20 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}\right )\right )}{\left (-1+e^{2 i (e+f x)}\right )^3 \left (1+e^{2 i (e+f x)}\right )^2}-\frac {3 \sqrt {b} (3 a+5 b) \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f (a+2 b+a \cos (2 e+2 f x))^{3/2}} \]
(Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2* I)*(e + f*x))]*Cos[e + f*x]^3*(((-I)*(4*a*(1 + E^((2*I)*(e + f*x)))^2*(1 - 4*E^((2*I)*(e + f*x)) + E^((4*I)*(e + f*x))) + b*(15 - 20*E^((2*I)*(e + f *x)) - 22*E^((4*I)*(e + f*x)) - 20*E^((6*I)*(e + f*x)) + 15*E^((8*I)*(e + f*x)))))/((-1 + E^((2*I)*(e + f*x)))^3*(1 + E^((2*I)*(e + f*x)))^2) - (3*S qrt[b]*(3*a + 5*b)*Log[(-4*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I)*Sq rt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2* I)*(e + f*x)))])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)) )^2])*(a + b*Sec[e + f*x]^2)^(3/2))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3 /2))
Time = 0.31 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4620, 359, 247, 211, 224, 219}
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.
| \(\displaystyle \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\sin (e+f x)^4}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \cot ^4(e+f x) \left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {(3 a+5 b) \int \cot ^2(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{3 (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\frac {(3 a+5 b) \left (3 b \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {(3 a+5 b) \left (3 b \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {(3 a+5 b) \left (3 b \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {(3 a+5 b) \left (3 b \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\) |
(-1/3*(Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(5/2))/(a + b) + ((3*a + 5*b)*(-(Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2)) + 3*b*(((a + b)*Arc Tanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*Sqrt[b]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2)))/(3*(a + b)))/f
3.1.91.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(1140\) vs. \(2(152)=304\).
Time = 10.65 (sec) , antiderivative size = 1141, normalized size of antiderivative = 6.63
-1/12/f/b*(15*cos(f*x+e)^3*sin(f*x+e)*b^(5/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1 +cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))+15*cos(f*x+e)^3*sin(f *x+e)*b^(5/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*c os(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a +a+b)/(sin(f*x+e)-1))-15*cos(f*x+e)^2*sin(f*x+e)*b^(5/2)*ln(4*(((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+ e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))-15*cos(f*x +e)^2*sin(f*x+e)*b^(5/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-s in(f*x+e)*a+a+b)/(sin(f*x+e)-1))+9*cos(f*x+e)^3*sin(f*x+e)*b^(3/2)*a*ln(4* (((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1) )+9*cos(f*x+e)^3*sin(f*x+e)*b^(3/2)*a*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f* x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e ))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))-9*cos(f*x+e)^2*sin(f*x+e)*b^ (3/2)*a*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+ e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/( sin(f*x+e)+1))-9*cos(f*x+e)^2*sin(f*x+e)*b^(3/2)*a*ln(-4*(((b+a*cos(f*x+e) ^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)...
Time = 1.50 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.74 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left ({\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (4 \, a + 15 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a + 10 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}, \frac {3 \, {\left ({\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (4 \, a + 15 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a + 10 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}\right ] \]
[1/24*(3*((3*a + 5*b)*cos(f*x + e)^3 - (3*a + 5*b)*cos(f*x + e))*sqrt(b)*l og(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4* ((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*((4*a + 15*b)*cos(f*x + e)^4 - 2*(3*a + 10*b)*cos(f*x + e)^2 + 3*b)*sqr t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((f*cos(f*x + e)^3 - f*cos(f*x + e))*sin(f*x + e)), 1/12*(3*((3*a + 5*b)*cos(f*x + e)^3 - (3*a + 5*b)*cos( f*x + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e)) *sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) - 2*((4*a + 15*b)*cos(f*x + e)^4 - 2*( 3*a + 10*b)*cos(f*x + e)^2 + 3*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) ^2))/((f*cos(f*x + e)^3 - f*cos(f*x + e))*sin(f*x + e))]
Timed out. \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.41 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {9 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \frac {6 \, a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{a + b} + 9 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \frac {6 \, b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{a + b} + 9 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b \tan \left (f x + e\right ) + \frac {6 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2} \tan \left (f x + e\right )}{a + b} - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )} - \frac {4 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}{{\left (a + b\right )} \tan \left (f x + e\right )} - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}}}{6 \, f} \]
1/6*(9*a*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 6*a*b^(3/2)*arc sinh(b*tan(f*x + e)/sqrt((a + b)*b))/(a + b) + 9*b^(3/2)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 6*b^(5/2)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) /(a + b) + 9*sqrt(b*tan(f*x + e)^2 + a + b)*b*tan(f*x + e) + 6*sqrt(b*tan( f*x + e)^2 + a + b)*b^2*tan(f*x + e)/(a + b) - 6*(b*tan(f*x + e)^2 + a + b )^(3/2)/tan(f*x + e) - 4*(b*tan(f*x + e)^2 + a + b)^(3/2)*b/((a + b)*tan(f *x + e)) - 2*(b*tan(f*x + e)^2 + a + b)^(5/2)/((a + b)*tan(f*x + e)^3))/f
\[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^4} \,d x \]