Integrand size = 16, antiderivative size = 98 \[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {3 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec ^2(c+d x)}}\right )}{8 d}+\frac {3 a^2 \sqrt {a \sec ^2(c+d x)} \tan (c+d x)}{8 d}+\frac {a \left (a \sec ^2(c+d x)\right )^{3/2} \tan (c+d x)}{4 d} \]
3/8*a^(5/2)*arctanh(a^(1/2)*tan(d*x+c)/(a*sec(d*x+c)^2)^(1/2))/d+1/4*a*(a* sec(d*x+c)^2)^(3/2)*tan(d*x+c)/d+3/8*a^2*(a*sec(d*x+c)^2)^(1/2)*tan(d*x+c) /d
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.60 \[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {a^2 \sqrt {a \sec ^2(c+d x)} \left (3 \text {arctanh}(\sin (c+d x)) \cos (c+d x)+\left (3+2 \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{8 d} \]
(a^2*Sqrt[a*Sec[c + d*x]^2]*(3*ArcTanh[Sin[c + d*x]]*Cos[c + d*x] + (3 + 2 *Sec[c + d*x]^2)*Tan[c + d*x]))/(8*d)
Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4140, 3042, 4610, 211, 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.
\(\displaystyle \int \left (a \tan ^2(c+d x)+a\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \tan (c+d x)^2+a\right )^{5/2}dx\) |
\(\Big \downarrow \) 4140 |
\(\displaystyle \int \left (a \sec ^2(c+d x)\right )^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sec (c+d x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle \frac {a \int \left (a \tan ^2(c+d x)+a\right )^{3/2}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {a \left (\frac {3}{4} a \int \sqrt {a \tan ^2(c+d x)+a}d\tan (c+d x)+\frac {1}{4} \tan (c+d x) \left (a \tan ^2(c+d x)+a\right )^{3/2}\right )}{d}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a \tan ^2(c+d x)+a}}d\tan (c+d x)+\frac {1}{2} \tan (c+d x) \sqrt {a \tan ^2(c+d x)+a}\right )+\frac {1}{4} \tan (c+d x) \left (a \tan ^2(c+d x)+a\right )^{3/2}\right )}{d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {a \tan ^2(c+d x)}{a \tan ^2(c+d x)+a}}d\frac {\tan (c+d x)}{\sqrt {a \tan ^2(c+d x)+a}}+\frac {1}{2} \tan (c+d x) \sqrt {a \tan ^2(c+d x)+a}\right )+\frac {1}{4} \tan (c+d x) \left (a \tan ^2(c+d x)+a\right )^{3/2}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a \left (\frac {3}{4} a \left (\frac {1}{2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \tan ^2(c+d x)+a}}\right )+\frac {1}{2} \tan (c+d x) \sqrt {a \tan ^2(c+d x)+a}\right )+\frac {1}{4} \tan (c+d x) \left (a \tan ^2(c+d x)+a\right )^{3/2}\right )}{d}\) |
(a*((Tan[c + d*x]*(a + a*Tan[c + d*x]^2)^(3/2))/4 + (3*a*((Sqrt[a]*ArcTanh [(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Tan[c + d*x]^2]])/2 + (Tan[c + d*x]*Sqr t[a + a*Tan[c + d*x]^2])/2))/4))/d
3.3.72.3.1 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[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*sec[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a, b]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFac tors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] && !IntegerQ[p]
Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {a \tan \left (d x +c \right ) \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}{4 d}+\frac {3 a^{2} \tan \left (d x +c \right ) \sqrt {a +a \tan \left (d x +c \right )^{2}}}{8 d}+\frac {3 a^{\frac {5}{2}} \ln \left (\sqrt {a}\, \tan \left (d x +c \right )+\sqrt {a +a \tan \left (d x +c \right )^{2}}\right )}{8 d}\) | \(90\) |
default | \(\frac {a \tan \left (d x +c \right ) \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}{4 d}+\frac {3 a^{2} \tan \left (d x +c \right ) \sqrt {a +a \tan \left (d x +c \right )^{2}}}{8 d}+\frac {3 a^{\frac {5}{2}} \ln \left (\sqrt {a}\, \tan \left (d x +c \right )+\sqrt {a +a \tan \left (d x +c \right )^{2}}\right )}{8 d}\) | \(90\) |
risch | \(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 i \left (d x +c \right )}+11 \,{\mathrm e}^{4 i \left (d x +c \right )}-11 \,{\mathrm e}^{2 i \left (d x +c \right )}-3\right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i d x}+i {\mathrm e}^{-i c}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, a^{2} \cos \left (d x +c \right )}{4 d}-\frac {3 \ln \left ({\mathrm e}^{i d x}-i {\mathrm e}^{-i c}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, a^{2} \cos \left (d x +c \right )}{4 d}\) | \(197\) |
1/4/d*a*tan(d*x+c)*(a+a*tan(d*x+c)^2)^(3/2)+3/8/d*a^2*tan(d*x+c)*(a+a*tan( d*x+c)^2)^(1/2)+3/8/d*a^(5/2)*ln(a^(1/2)*tan(d*x+c)+(a+a*tan(d*x+c)^2)^(1/ 2))
Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93 \[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {3 \, a^{\frac {5}{2}} \log \left (2 \, a \tan \left (d x + c\right )^{2} + 2 \, \sqrt {a \tan \left (d x + c\right )^{2} + a} \sqrt {a} \tan \left (d x + c\right ) + a\right ) + 2 \, {\left (2 \, a^{2} \tan \left (d x + c\right )^{3} + 5 \, a^{2} \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{16 \, d} \]
1/16*(3*a^(5/2)*log(2*a*tan(d*x + c)^2 + 2*sqrt(a*tan(d*x + c)^2 + a)*sqrt (a)*tan(d*x + c) + a) + 2*(2*a^2*tan(d*x + c)^3 + 5*a^2*tan(d*x + c))*sqrt (a*tan(d*x + c)^2 + a))/d
\[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\int \left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {5}{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1769 vs. \(2 (82) = 164\).
Time = 0.58 (sec) , antiderivative size = 1769, normalized size of antiderivative = 18.05 \[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\text {Too large to display} \]
1/16*(176*a^2*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 48*a^2*cos(d*x + c)*sin( 2*d*x + 2*c) - 48*a^2*cos(2*d*x + 2*c)*sin(d*x + c) - 12*a^2*sin(d*x + c) + 4*(3*a^2*sin(7*d*x + 7*c) + 11*a^2*sin(5*d*x + 5*c) - 11*a^2*sin(3*d*x + 3*c) - 3*a^2*sin(d*x + c))*cos(8*d*x + 8*c) - 24*(2*a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*cos(7*d*x + 7*c) + 16*( 11*a^2*sin(5*d*x + 5*c) - 11*a^2*sin(3*d*x + 3*c) - 3*a^2*sin(d*x + c))*co s(6*d*x + 6*c) - 88*(3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*cos( 5*d*x + 5*c) - 24*(11*a^2*sin(3*d*x + 3*c) + 3*a^2*sin(d*x + c))*cos(4*d*x + 4*c) + 3*(a^2*cos(8*d*x + 8*c)^2 + 16*a^2*cos(6*d*x + 6*c)^2 + 36*a^2*c os(4*d*x + 4*c)^2 + 16*a^2*cos(2*d*x + 2*c)^2 + a^2*sin(8*d*x + 8*c)^2 + 1 6*a^2*sin(6*d*x + 6*c)^2 + 36*a^2*sin(4*d*x + 4*c)^2 + 48*a^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*a^2*sin(2*d*x + 2*c)^2 + 8*a^2*cos(2*d*x + 2*c) + a^2 + 2*(4*a^2*cos(6*d*x + 6*c) + 6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos(2* d*x + 2*c) + a^2)*cos(8*d*x + 8*c) + 8*(6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos (2*d*x + 2*c) + a^2)*cos(6*d*x + 6*c) + 12*(4*a^2*cos(2*d*x + 2*c) + a^2)* cos(4*d*x + 4*c) + 4*(2*a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 2* a^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*a^2*sin(4*d*x + 4*c) + 2*a^ 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - 3*(a^2*cos(8*d*x + 8*c)^2 + 16*a^2*cos(6*d*x + 6*c )^2 + 36*a^2*cos(4*d*x + 4*c)^2 + 16*a^2*cos(2*d*x + 2*c)^2 + a^2*sin(8...
Leaf count of result is larger than twice the leaf count of optimal. 7084 vs. \(2 (82) = 164\).
Time = 3.58 (sec) , antiderivative size = 7084, normalized size of antiderivative = 72.29 \[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\text {Too large to display} \]
1/8*(3*(a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2 *c)^3 - tan(1/2*d*x)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan( 1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x)*tan(1/2*c) + 1)*tan(1/2*c) - a^(5 /2)*sgn(tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan( 1/2*d*x)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - t an(1/2*c)^4 - 4*tan(1/2*d*x)*tan(1/2*c) + 1))*log(abs(-tan(1/2*d*x)*tan(1/ 2*c) + tan(1/2*d*x) + tan(1/2*c) + 1))/(tan(1/2*c) - 1) - 3*(a^(5/2)*sgn(t an(1/2*d*x)^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*d*x)^ 4 - 4*tan(1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c) ^4 - 4*tan(1/2*d*x)*tan(1/2*c) + 1)*tan(1/2*c) + a^(5/2)*sgn(tan(1/2*d*x)^ 4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*d*x)^4 - 4*tan(1/ 2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1 /2*d*x)*tan(1/2*c) + 1))*log(abs(-tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) + 1))/(tan(1/2*c) + 1) - 2*(5*a^(5/2)*sgn(tan(1/2*d*x)^4*tan(1 /2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*d*x)^4 - 4*tan(1/2*d*x)^ 3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan(1/2*d*x) *tan(1/2*c) + 1)*tan(1/2*d*x)^7*tan(1/2*c)^16 + 34*a^(5/2)*sgn(tan(1/2*d*x )^4*tan(1/2*c)^4 - 4*tan(1/2*d*x)^3*tan(1/2*c)^3 - tan(1/2*d*x)^4 - 4*tan( 1/2*d*x)^3*tan(1/2*c) - 4*tan(1/2*d*x)*tan(1/2*c)^3 - tan(1/2*c)^4 - 4*tan (1/2*d*x)*tan(1/2*c) + 1)*tan(1/2*d*x)^7*tan(1/2*c)^14 - 6*a^(5/2)*sgn(...
Timed out. \[ \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx=\int {\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}^{5/2} \,d x \]