Integrand size = 41, antiderivative size = 170 \[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\frac {2 C \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {2 (A-C) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (1+2 m),\frac {1}{4} (5+2 m),-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {2 B \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (3+2 m) \sqrt {b \tan (c+d x)}} \]
2*C*tan(d*x+c)^(1+m)/d/(1+2*m)/(b*tan(d*x+c))^(1/2)+2*(A-C)*hypergeom([1, 1/4+1/2*m],[5/4+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(1+m)/d/(1+2*m)/(b*tan(d* x+c))^(1/2)+2*B*hypergeom([1, 3/4+1/2*m],[7/4+1/2*m],-tan(d*x+c)^2)*tan(d* x+c)^(2+m)/d/(3+2*m)/(b*tan(d*x+c))^(1/2)
Time = 0.53 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.78 \[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\frac {2 \tan ^{1+m}(c+d x) \left (C (3+2 m)+(A-C) (3+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (1+2 m),\frac {1}{4} (5+2 m),-\tan ^2(c+d x)\right )+B (1+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(c+d x)\right ) \tan (c+d x)\right )}{d (1+2 m) (3+2 m) \sqrt {b \tan (c+d x)}} \]
(2*Tan[c + d*x]^(1 + m)*(C*(3 + 2*m) + (A - C)*(3 + 2*m)*Hypergeometric2F1 [1, (1 + 2*m)/4, (5 + 2*m)/4, -Tan[c + d*x]^2] + B*(1 + 2*m)*Hypergeometri c2F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -Tan[c + d*x]^2]*Tan[c + d*x]))/(d*(1 + 2*m)*(3 + 2*m)*Sqrt[b*Tan[c + d*x]])
Time = 0.54 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2034, 3042, 4113, 3042, 4021, 3042, 3957, 278}
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.
| \(\displaystyle \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \int \tan ^{m-\frac {1}{2}}(c+d x) \left (C \tan ^2(c+d x)+B \tan (c+d x)+A\right )dx}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \int \tan (c+d x)^{m-\frac {1}{2}} \left (C \tan (c+d x)^2+B \tan (c+d x)+A\right )dx}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \left (\int \tan ^{m-\frac {1}{2}}(c+d x) (A-C+B \tan (c+d x))dx+\frac {2 C \tan ^{m+\frac {1}{2}}(c+d x)}{d (2 m+1)}\right )}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \left (\int \tan (c+d x)^{m-\frac {1}{2}} (A-C+B \tan (c+d x))dx+\frac {2 C \tan ^{m+\frac {1}{2}}(c+d x)}{d (2 m+1)}\right )}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4021 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \left ((A-C) \int \tan ^{m-\frac {1}{2}}(c+d x)dx+B \int \tan ^{m+\frac {1}{2}}(c+d x)dx+\frac {2 C \tan ^{m+\frac {1}{2}}(c+d x)}{d (2 m+1)}\right )}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \left ((A-C) \int \tan (c+d x)^{m-\frac {1}{2}}dx+B \int \tan (c+d x)^{m+\frac {1}{2}}dx+\frac {2 C \tan ^{m+\frac {1}{2}}(c+d x)}{d (2 m+1)}\right )}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \left (\frac {(A-C) \int \frac {\tan ^{m-\frac {1}{2}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {B \int \frac {\tan ^{m+\frac {1}{2}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {2 C \tan ^{m+\frac {1}{2}}(c+d x)}{d (2 m+1)}\right )}{\sqrt {b \tan (c+d x)}}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {\sqrt {\tan (c+d x)} \left (\frac {2 (A-C) \tan ^{m+\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2 m+1),\frac {1}{4} (2 m+5),-\tan ^2(c+d x)\right )}{d (2 m+1)}+\frac {2 B \tan ^{m+\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2 m+3),\frac {1}{4} (2 m+7),-\tan ^2(c+d x)\right )}{d (2 m+3)}+\frac {2 C \tan ^{m+\frac {1}{2}}(c+d x)}{d (2 m+1)}\right )}{\sqrt {b \tan (c+d x)}}\) |
(Sqrt[Tan[c + d*x]]*((2*C*Tan[c + d*x]^(1/2 + m))/(d*(1 + 2*m)) + (2*(A - C)*Hypergeometric2F1[1, (1 + 2*m)/4, (5 + 2*m)/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(1/2 + m))/(d*(1 + 2*m)) + (2*B*Hypergeometric2F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(3/2 + m))/(d*(3 + 2*m))))/Sqrt[b *Tan[c + d*x]]
3.1.48.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart [n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n])) Int[(a*v)^(m + n )*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[m + n]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Tan[e + f*x])^m, x], x] + Simp[d/b Int [(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ[c^ 2 + d^2, 0] && !IntegerQ[2*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]
\[\int \frac {\tan \left (d x +c \right )^{m} \left (A +B \tan \left (d x +c \right )+C \tan \left (d x +c \right )^{2}\right )}{\sqrt {b \tan \left (d x +c \right )}}d x\]
\[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\int { \frac {{\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{\sqrt {b \tan \left (d x + c\right )}} \,d x } \]
integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(b*tan(d*x+c))^(1/2 ),x, algorithm="fricas")
integral((C*tan(d*x + c)^2 + B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c))*tan( d*x + c)^m/(b*tan(d*x + c)), x)
\[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )} + C \tan ^{2}{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\sqrt {b \tan {\left (c + d x \right )}}}\, dx \]
Timed out. \[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\text {Timed out} \]
integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(b*tan(d*x+c))^(1/2 ),x, algorithm="maxima")
Timed out. \[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\text {Timed out} \]
integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(b*tan(d*x+c))^(1/2 ),x, algorithm="giac")
Timed out. \[ \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (C\,{\mathrm {tan}\left (c+d\,x\right )}^2+B\,\mathrm {tan}\left (c+d\,x\right )+A\right )}{\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]