Integrand size = 25, antiderivative size = 148 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=-\frac {\sqrt {a+b} (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f} \]
1/6*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)/f+1/2*sec(f*x+e)^2*(a+b*sin(f *x+e)^2)^(5/2)/(a+b)/f-1/2*(2*a+5*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b )^(1/2))*(a+b)^(1/2)/f+1/2*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.54 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {3 \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}+(2 a+5 b) \left (-3 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+3 b+b \sin ^2(e+f x)\right )\right )}{6 (a+b) f} \]
(3*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) + (2*a + 5*b)*(-3*(a + b)^( 3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqrt[a + b*Sin[e + f*x]^2]*(4*a + 3*b + b*Sin[e + f*x]^2)))/(6*(a + b)*f)
Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 87, 60, 60, 73, 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.
\(\displaystyle \int \tan ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^3 \left (a+b \sin (e+f x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\sin ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )^{3/2}}{\left (1-\sin ^2(e+f x)\right )^2}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+5 b) \int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)}{2 (a+b)}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+5 b) \left ((a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+5 b) \left ((a+b) \left ((a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+5 b) \left ((a+b) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+5 b) \left ((a+b) \left (2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{2 f}\) |
((a + b*Sin[e + f*x]^2)^(5/2)/((a + b)*(1 - Sin[e + f*x]^2)) - ((2*a + 5*b )*((-2*(a + b*Sin[e + f*x]^2)^(3/2))/3 + (a + b)*(2*Sqrt[a + b]*ArcTanh[Sq rt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - 2*Sqrt[a + b*Sin[e + f*x]^2])))/(2 *(a + b)))/(2*f)
3.5.100.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(128)=256\).
Time = 1.78 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.88
method | result | size |
default | \(\frac {-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (b \left (\cos ^{2}\left (f x +e \right )\right )+2 a -b \right )}{3}+2 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}+2 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-\frac {\left (\frac {1}{2} a^{2}+2 a b +\frac {3}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\sqrt {a +b}}-\frac {\left (\frac {1}{2} a^{2}+2 a b +\frac {3}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\sqrt {a +b}}+\left (\frac {1}{4} a^{2}+\frac {1}{2} a b +\frac {1}{4} b^{2}\right ) \left (-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{\left (a +b \right ) \left (\sin \left (f x +e \right )-1\right )}+\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )+\left (-\frac {1}{4} a^{2}-\frac {1}{2} a b -\frac {1}{4} b^{2}\right ) \left (-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{\left (a +b \right ) \left (1+\sin \left (f x +e \right )\right )}-\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )}{f}\) | \(426\) |
(-1/3*(a+b-b*cos(f*x+e)^2)^(1/2)*(b*cos(f*x+e)^2+2*a-b)+2*b*(a+b*sin(f*x+e )^2)^(1/2)+2*a*(a+b*sin(f*x+e)^2)^(1/2)-(1/2*a^2+2*a*b+3/2*b^2)/(a+b)^(1/2 )*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin( f*x+e)))-(1/2*a^2+2*a*b+3/2*b^2)/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos( f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1))+(1/4*a^2+1/2*a*b+1/4*b ^2)*(-1/(a+b)/(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)^(1/2)+b/(a+b)^(3/2)*ln(( 2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1 )))+(-1/4*a^2-1/2*a*b-1/4*b^2)*(-1/(a+b)/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^ 2)^(1/2)-b/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*si n(f*x+e)+2*a)/(1+sin(f*x+e)))))/f
Time = 0.53 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.79 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\left [\frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, f \cos \left (f x + e\right )^{2}}, \frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{2} - {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f \cos \left (f x + e\right )^{2}}\right ] \]
[1/12*(3*(2*a + 5*b)*sqrt(a + b)*cos(f*x + e)^2*log((b*cos(f*x + e)^2 + 2* sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*(2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b)*sqrt(-b *cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2), 1/6*(3*(2*a + 5*b)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^2 - (2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b)*sq rt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2)]
Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.14 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {4 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2} + 12 \, {\left (a b^{2} + 2 \, b^{3}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {3 \, {\left (2 \, a^{2} b^{2} + 7 \, a b^{3} + 5 \, b^{4}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}} - \frac {6 \, {\left (a b^{3} + b^{4}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}{b \sin \left (f x + e\right )^{2} - b}}{12 \, b^{2} f} \]
1/12*(4*(b*sin(f*x + e)^2 + a)^(3/2)*b^2 + 12*(a*b^2 + 2*b^3)*sqrt(b*sin(f *x + e)^2 + a) + 3*(2*a^2*b^2 + 7*a*b^3 + 5*b^4)*log((sqrt(b*sin(f*x + e)^ 2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/sqrt(a + b) - 6*(a*b^3 + b^4)*sqrt(b*sin(f*x + e)^2 + a)/(b*sin(f*x + e)^2 - b))/( b^2*f)
Exception generated. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]