Integrand size = 23, antiderivative size = 109 \[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {\sqrt {a-b} (2 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}-\frac {(a-b) \sin (c+d x)}{2 a b d \left (a-(a-b) \sin ^2(c+d x)\right )} \]
arctanh(sin(d*x+c))/b^2/d-1/2*(a-b)*sin(d*x+c)/a/b/d/(a-(a-b)*sin(d*x+c)^2 )-1/2*(2*a+b)*arctanh(sin(d*x+c)*(a-b)^(1/2)/a^(1/2))*(a-b)^(1/2)/a^(3/2)/ b^2/d
Time = 0.90 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.75 \[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sqrt {a-b} (2 a+b) \log \left (\sqrt {a}-\sqrt {a-b} \sin (c+d x)\right )}{a^{3/2}}+\frac {\left (-2 a^2+a b+b^2\right ) \log \left (\sqrt {a}+\sqrt {a-b} \sin (c+d x)\right )}{a^{3/2} \sqrt {a-b}}+\frac {4 b (-a+b) \sin (c+d x)}{a (a+b+(a-b) \cos (2 (c+d x)))}}{4 b^2 d} \]
(-4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*Log[Cos[(c + d*x)/2] + Si n[(c + d*x)/2]] + (Sqrt[a - b]*(2*a + b)*Log[Sqrt[a] - Sqrt[a - b]*Sin[c + d*x]])/a^(3/2) + ((-2*a^2 + a*b + b^2)*Log[Sqrt[a] + Sqrt[a - b]*Sin[c + d*x]])/(a^(3/2)*Sqrt[a - b]) + (4*b*(-a + b)*Sin[c + d*x])/(a*(a + b + (a - b)*Cos[2*(c + d*x)])))/(4*b^2*d)
Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4159, 316, 25, 397, 219, 221}
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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^5}{\left (a+b \tan (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )^2}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {-\frac {\int -\frac {(a-b) \sin ^2(c+d x)+a+b}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}d\sin (c+d x)}{2 a b}-\frac {(a-b) \sin (c+d x)}{2 a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {(a-b) \sin ^2(c+d x)+a+b}{\left (1-\sin ^2(c+d x)\right ) \left (a-(a-b) \sin ^2(c+d x)\right )}d\sin (c+d x)}{2 a b}-\frac {(a-b) \sin (c+d x)}{2 a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {2 a \int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)}{b}-\frac {(a-b) (2 a+b) \int \frac {1}{a-(a-b) \sin ^2(c+d x)}d\sin (c+d x)}{b}}{2 a b}-\frac {(a-b) \sin (c+d x)}{2 a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {2 a \text {arctanh}(\sin (c+d x))}{b}-\frac {(a-b) (2 a+b) \int \frac {1}{a-(a-b) \sin ^2(c+d x)}d\sin (c+d x)}{b}}{2 a b}-\frac {(a-b) \sin (c+d x)}{2 a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {2 a \text {arctanh}(\sin (c+d x))}{b}-\frac {\sqrt {a-b} (2 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b}}{2 a b}-\frac {(a-b) \sin (c+d x)}{2 a b \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
(((2*a*ArcTanh[Sin[c + d*x]])/b - (Sqrt[a - b]*(2*a + b)*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*b))/(2*a*b) - ((a - b)*Sin[c + d*x])/ (2*a*b*(a - (a - b)*Sin[c + d*x]^2)))/d
3.5.63.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 17.67 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 b^{2}}+\frac {\left (a -b \right ) \left (\frac {b \sin \left (d x +c \right )}{2 a \left (\sin \left (d x +c \right )^{2} a -b \sin \left (d x +c \right )^{2}-a \right )}-\frac {\left (2 a +b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{b^{2}}+\frac {\ln \left (\sin \left (d x +c \right )+1\right )}{2 b^{2}}}{d}\) | \(124\) |
| default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 b^{2}}+\frac {\left (a -b \right ) \left (\frac {b \sin \left (d x +c \right )}{2 a \left (\sin \left (d x +c \right )^{2} a -b \sin \left (d x +c \right )^{2}-a \right )}-\frac {\left (2 a +b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{b^{2}}+\frac {\ln \left (\sin \left (d x +c \right )+1\right )}{2 b^{2}}}{d}\) | \(124\) |
| risch | \(\frac {i \left (a -b \right ) \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{a b d \left (a \,{\mathrm e}^{4 i \left (d x +c \right )}-b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \,b^{2}}+\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{2 a d \,b^{2}}+\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a^{2} d b}-\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{2 a d \,b^{2}}-\frac {\sqrt {a \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {a \left (a -b \right )}\, {\mathrm e}^{i \left (d x +c \right )}}{a -b}-1\right )}{4 a^{2} d b}\) | \(375\) |
1/d*(-1/2/b^2*ln(sin(d*x+c)-1)+(a-b)/b^2*(1/2*b/a*sin(d*x+c)/(sin(d*x+c)^2 *a-b*sin(d*x+c)^2-a)-1/2*(2*a+b)/a/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c )/(a*(a-b))^(1/2)))+1/2/b^2*ln(sin(d*x+c)+1))
Time = 0.37 (sec) , antiderivative size = 407, normalized size of antiderivative = 3.73 \[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [\frac {{\left ({\left (2 \, a^{2} - a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a - b}{a}} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a \sqrt {\frac {a - b}{a}} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) + 2 \, {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a b - b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a b^{3} d + {\left (a^{2} b^{2} - a b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}, \frac {{\left ({\left (2 \, a^{2} - a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a - b}{a}} \arctan \left (\sqrt {-\frac {a - b}{a}} \sin \left (d x + c\right )\right ) + {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a b - b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a b^{3} d + {\left (a^{2} b^{2} - a b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
[1/4*(((2*a^2 - a*b - b^2)*cos(d*x + c)^2 + 2*a*b + b^2)*sqrt((a - b)/a)*l og(-((a - b)*cos(d*x + c)^2 + 2*a*sqrt((a - b)/a)*sin(d*x + c) - 2*a + b)/ ((a - b)*cos(d*x + c)^2 + b)) + 2*((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log(s in(d*x + c) + 1) - 2*((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log(-sin(d*x + c) + 1) - 2*(a*b - b^2)*sin(d*x + c))/(a*b^3*d + (a^2*b^2 - a*b^3)*d*cos(d*x + c)^2), 1/2*(((2*a^2 - a*b - b^2)*cos(d*x + c)^2 + 2*a*b + b^2)*sqrt(-(a - b)/a)*arctan(sqrt(-(a - b)/a)*sin(d*x + c)) + ((a^2 - a*b)*cos(d*x + c)^ 2 + a*b)*log(sin(d*x + c) + 1) - ((a^2 - a*b)*cos(d*x + c)^2 + a*b)*log(-s in(d*x + c) + 1) - (a*b - b^2)*sin(d*x + c))/(a*b^3*d + (a^2*b^2 - a*b^3)* d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Time = 0.62 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{b^{2}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{b^{2}} - \frac {{\left (2 \, a^{2} - a b - b^{2}\right )} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} a b^{2}} + \frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )} a b}}{2 \, d} \]
1/2*(log(abs(sin(d*x + c) + 1))/b^2 - log(abs(sin(d*x + c) - 1))/b^2 - (2* a^2 - a*b - b^2)*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b ))/(sqrt(-a^2 + a*b)*a*b^2) + (a*sin(d*x + c) - b*sin(d*x + c))/((a*sin(d* x + c)^2 - b*sin(d*x + c)^2 - a)*a*b))/d
Time = 14.57 (sec) , antiderivative size = 946, normalized size of antiderivative = 8.68 \[ \int \frac {\sec ^5(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
((a^2*atan((sin(c/2 + (d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d *x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*(b - a)^(1/2)*1i - a ^(5/2)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*2i + (b^2*atan((sin(c/ 2 + (d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*c os(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*(b - a)^(1/2)*1i)/2 - a^(3/2)*b*atanh( sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*2i - a^(1/2)*b^2*sin(c + d*x)*1i + a^2*atan((a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)) )*(b - a)^(1/2)*1i + (b^2*atan((a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + ( d*x)/2)*(b - a)^(1/2)))*(b - a)^(1/2)*1i)/2 - a^(5/2)*atanh(sin(c/2 + (d*x )/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*2i + a^(3/2)*b*sin(c + d*x)*1i + a^2*atan((a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2) ))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i - (b^2*atan((a^(1/2)*sin(c/2 + (d*x)/ 2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1 i)/2 + (a*b*atan((sin(c/2 + (d*x)/2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos (c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*(b - a)^(1/2)* 3i)/2 + (a*b*atan((a^(1/2)*sin(c/2 + (d*x)/2))/(2*cos(c/2 + (d*x)/2)*(b - a)^(1/2)))*(b - a)^(1/2)*3i)/2 + a^2*atan((sin(c/2 + (d*x)/2)*(a - 2*b + 2 *a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 + (d*x)/2)^3*(b - a)^(1/2)))*cos(2*c + 2*d*x)*(b - a)^(1/2)*1i - (b^2*atan((sin(c/2 + (d*x)/ 2)*(a - 2*b + 2*a*cos(c + d*x) - 2*b*cos(c + d*x)))/(2*a^(1/2)*cos(c/2 ...