Integrand size = 45, antiderivative size = 201 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {(i a+b) (A-i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {(i a-b) (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}+\frac {2 (b c-a d) \left (c^2 C-B c d+A d^2\right )}{d^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 b C \sqrt {c+d \tan (e+f x)}}{d^2 f} \]
-(I*a+b)*(A-I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^( 3/2)/f+(I*a-b)*(A+I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+ I*d)^(3/2)/f+2*(-a*d+b*c)*(A*d^2-B*c*d+C*c^2)/d^2/(c^2+d^2)/f/(c+d*tan(f*x +e))^(1/2)+2*b*C*(c+d*tan(f*x+e))^(1/2)/d^2/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.78 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {(A b+a B-b C) \left (-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )-\frac {2 (-2 b c C+b B d+2 a C d)}{d \sqrt {c+d \tan (e+f x)}}+\frac {(A b c+a B c-b c C-a A d+b B d+a C d) \left ((-i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )+(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right )}{\left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {2 C (a+b \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}}}{d f} \]
Integrate[((a + b*Tan[e + f*x])*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/( c + d*Tan[e + f*x])^(3/2),x]
((A*b + a*B - b*C)*(((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]) /Sqrt[c - I*d] + (I*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[ c + I*d]) - (2*(-2*b*c*C + b*B*d + 2*a*C*d))/(d*Sqrt[c + d*Tan[e + f*x]]) + ((A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d)*(((-I)*c + d)*Hypergeom etric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] + (I*c + d)*Hyperge ometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]))/((c^2 + d^2)*Sq rt[c + d*Tan[e + f*x]]) + (2*C*(a + b*Tan[e + f*x]))/Sqrt[c + d*Tan[e + f* x]])/(d*f)
Time = 1.16 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4118, 3042, 4113, 3042, 4022, 3042, 4020, 25, 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.
\(\displaystyle \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )}{(c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4118 |
\(\displaystyle \frac {\int \frac {b C \left (c^2+d^2\right ) \tan ^2(e+f x)+d (A b c+a B c-b C c-a A d+b B d+a C d) \tan (e+f x)+a d (A c-C c+B d)+b \left (C c^2-B d c+A d^2\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}+\frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {b C \left (c^2+d^2\right ) \tan (e+f x)^2+d (A b c+a B c-b C c-a A d+b B d+a C d) \tan (e+f x)+a d (A c-C c+B d)+b \left (C c^2-B d c+A d^2\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}+\frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {\int \frac {d (a (A c-C c+B d)-b (B c-(A-C) d))+d (A b c+a B c-b C c-a A d+b B d+a C d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}+\frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {d (a (A c-C c+B d)-b (B c-(A-C) d))+d (A b c+a B c-b C c-a A d+b B d+a C d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}+\frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} d (a+i b) (c-i d) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} d (a-i b) (c+i d) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} d (a+i b) (c-i d) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} d (a-i b) (c+i d) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i d (a-i b) (c+i d) (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i d (a+i b) (c-i d) (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {i d (a-i b) (c+i d) (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i d (a+i b) (c-i d) (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d) (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {(a-i b) (c+i d) (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (b c-a d) \left (A d^2-B c d+c^2 C\right )}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {d (a-i b) (c+i d) (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {d (a+i b) (c-i d) (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {2 b C \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\) |
Int[((a + b*Tan[e + f*x])*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(c + d* Tan[e + f*x])^(3/2),x]
(2*(b*c - a*d)*(c^2*C - B*c*d + A*d^2))/(d^2*(c^2 + d^2)*f*Sqrt[c + d*Tan[ e + f*x]]) + (((a - I*b)*(A - I*B - C)*(c + I*d)*d*ArcTan[Tan[e + f*x]/Sqr t[c - I*d]])/(Sqrt[c - I*d]*f) + ((a + I*b)*(A + I*B - C)*(c - I*d)*d*ArcT an[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) + (2*b*C*(c^2 + d^2)*Sqr t[c + d*Tan[e + f*x]])/(d*f))/(d*(c^2 + d^2))
3.2.18.3.1 Defintions of rubi rules used
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_)^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_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_. )*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* (c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) *Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n , -1]
Leaf count of result is larger than twice the leaf count of optimal. \(7395\) vs. \(2(177)=354\).
Time = 0.16 (sec) , antiderivative size = 7396, normalized size of antiderivative = 36.80
method | result | size |
parts | \(\text {Expression too large to display}\) | \(7396\) |
derivativedivides | \(\text {Expression too large to display}\) | \(23472\) |
default | \(\text {Expression too large to display}\) | \(23472\) |
int((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(3/2 ),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 31879 vs. \(2 (170) = 340\).
Time = 61.21 (sec) , antiderivative size = 31879, normalized size of antiderivative = 158.60 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) )^(3/2),x, algorithm="fricas")
\[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right ) \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Integral((a + b*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(c + d*tan(e + f*x))**(3/2), x)
Timed out. \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) )^(3/2),x, algorithm="maxima")
Timed out. \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
integrate((a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) )^(3/2),x, algorithm="giac")
Time = 38.60 (sec) , antiderivative size = 40542, normalized size of antiderivative = 201.70 \[ \int \frac {(a+b \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
int(((a + b*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(c + d* tan(e + f*x))^(3/2),x)
atan((((c + d*tan(e + f*x))^(1/2)*(16*A^2*a^2*d^10*f^3 - 16*B^2*a^2*d^10*f ^3 + 16*C^2*a^2*d^10*f^3 + 32*A^2*a^2*c^2*d^8*f^3 - 32*A^2*a^2*c^6*d^4*f^3 - 16*A^2*a^2*c^8*d^2*f^3 - 32*B^2*a^2*c^2*d^8*f^3 + 32*B^2*a^2*c^6*d^4*f^ 3 + 16*B^2*a^2*c^8*d^2*f^3 + 32*C^2*a^2*c^2*d^8*f^3 - 32*C^2*a^2*c^6*d^4*f ^3 - 16*C^2*a^2*c^8*d^2*f^3 - 32*A*C*a^2*d^10*f^3 - 64*A*B*a^2*c*d^9*f^3 + 64*B*C*a^2*c*d^9*f^3 - 192*A*B*a^2*c^3*d^7*f^3 - 192*A*B*a^2*c^5*d^5*f^3 - 64*A*B*a^2*c^7*d^3*f^3 - 64*A*C*a^2*c^2*d^8*f^3 + 64*A*C*a^2*c^6*d^4*f^3 + 32*A*C*a^2*c^8*d^2*f^3 + 192*B*C*a^2*c^3*d^7*f^3 + 192*B*C*a^2*c^5*d^5* f^3 + 64*B*C*a^2*c^7*d^3*f^3) - ((((8*A^2*a^2*c^3*f^2 - 8*B^2*a^2*c^3*f^2 + 8*C^2*a^2*c^3*f^2 - 16*A*B*a^2*d^3*f^2 - 16*A*C*a^2*c^3*f^2 + 16*B*C*a^2 *d^3*f^2 - 24*A^2*a^2*c*d^2*f^2 + 24*B^2*a^2*c*d^2*f^2 - 24*C^2*a^2*c*d^2* f^2 + 48*A*B*a^2*c^2*d*f^2 + 48*A*C*a^2*c*d^2*f^2 - 48*B*C*a^2*c^2*d*f^2)^ 2/4 - (16*c^6*f^4 + 16*d^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4)*(A^4*a^4 + B^4*a^4 + C^4*a^4 - 4*A*C^3*a^4 - 4*A^3*C*a^4 + 2*A^2*B^2*a^4 + 6*A^2*C ^2*a^4 + 2*B^2*C^2*a^4 - 4*A*B^2*C*a^4))^(1/2) - 4*A^2*a^2*c^3*f^2 + 4*B^2 *a^2*c^3*f^2 - 4*C^2*a^2*c^3*f^2 + 8*A*B*a^2*d^3*f^2 + 8*A*C*a^2*c^3*f^2 - 8*B*C*a^2*d^3*f^2 + 12*A^2*a^2*c*d^2*f^2 - 12*B^2*a^2*c*d^2*f^2 + 12*C^2* a^2*c*d^2*f^2 - 24*A*B*a^2*c^2*d*f^2 - 24*A*C*a^2*c*d^2*f^2 + 24*B*C*a^2*c ^2*d*f^2)/(16*(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4)))^(1/2)* ((c + d*tan(e + f*x))^(1/2)*((((8*A^2*a^2*c^3*f^2 - 8*B^2*a^2*c^3*f^2 +...