Integrand size = 27, antiderivative size = 219 \[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\frac {3 a b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {a \left (a^2-3 b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+n p),\frac {1}{2} (3+n p),-\tan ^2(e+f x)\right ) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {b^3 \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (2+n p)}+\frac {b \left (3 a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2+n p),\frac {1}{2} (4+n p),-\tan ^2(e+f x)\right ) \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (2+n p)} \]
3*a*b^2*tan(f*x+e)*(c*(d*tan(f*x+e))^p)^n/f/(n*p+1)+a*(a^2-3*b^2)*hypergeo m([1, 1/2*n*p+1/2],[1/2*n*p+3/2],-tan(f*x+e)^2)*tan(f*x+e)*(c*(d*tan(f*x+e ))^p)^n/f/(n*p+1)+b^3*tan(f*x+e)^2*(c*(d*tan(f*x+e))^p)^n/f/(n*p+2)+b*(3*a ^2-b^2)*hypergeom([1, 1/2*n*p+1],[1/2*n*p+2],-tan(f*x+e)^2)*tan(f*x+e)^2*( c*(d*tan(f*x+e))^p)^n/f/(n*p+2)
Time = 1.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.74 \[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n \left (a \left (a^2-3 b^2\right ) (2+n p) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+n p),\frac {1}{2} (3+n p),-\tan ^2(e+f x)\right )+b \left (\left (3 a^2-b^2\right ) (1+n p) \operatorname {Hypergeometric2F1}\left (1,1+\frac {n p}{2},2+\frac {n p}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)+b (3 a (2+n p)+b (1+n p) \tan (e+f x))\right )\right )}{f (1+n p) (2+n p)} \]
(Tan[e + f*x]*(c*(d*Tan[e + f*x])^p)^n*(a*(a^2 - 3*b^2)*(2 + n*p)*Hypergeo metric2F1[1, (1 + n*p)/2, (3 + n*p)/2, -Tan[e + f*x]^2] + b*((3*a^2 - b^2) *(1 + n*p)*Hypergeometric2F1[1, 1 + (n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2] *Tan[e + f*x] + b*(3*a*(2 + n*p) + b*(1 + n*p)*Tan[e + f*x]))))/(f*(1 + n* p)*(2 + n*p))
Time = 0.57 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4853, 2042, 559, 2333, 2009}
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.
| \(\displaystyle \int (a+b \tan (e+f x))^3 \left (c (d \tan (e+f x))^p\right )^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x))^3 \left (c (d \tan (e+f x))^p\right )^ndx\) |
| \(\Big \downarrow \) 4853 |
| \(\displaystyle \frac {\int \frac {\left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2042 |
| \(\displaystyle \frac {\tan ^{-n p}(e+f x) \left (c (d \tan (e+f x))^p\right )^n \int \frac {\tan ^{n p}(e+f x) (a+b \tan (e+f x))^3}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 559 |
| \(\displaystyle \frac {\tan ^{-n p}(e+f x) \left (c (d \tan (e+f x))^p\right )^n \left (\frac {\int \frac {\tan ^{n p}(e+f x) \left ((n p+2) a^3+3 b^2 (n p+2) \tan ^2(e+f x) a+b \left (3 a^2-b^2\right ) (n p+2) \tan (e+f x)\right )}{\tan ^2(e+f x)+1}d\tan (e+f x)}{n p+2}+\frac {b^3 \tan ^{n p+2}(e+f x)}{n p+2}\right )}{f}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\tan ^{-n p}(e+f x) \left (c (d \tan (e+f x))^p\right )^n \left (\frac {\int \left (3 a b^2 (n p+2) \tan ^{n p}(e+f x)+\frac {\left (a \left (a^2-3 b^2\right ) (n p+2)+b \left (3 a^2-b^2\right ) \tan (e+f x) (n p+2)\right ) \tan ^{n p}(e+f x)}{\tan ^2(e+f x)+1}\right )d\tan (e+f x)}{n p+2}+\frac {b^3 \tan ^{n p+2}(e+f x)}{n p+2}\right )}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\tan ^{-n p}(e+f x) \left (c (d \tan (e+f x))^p\right )^n \left (\frac {\frac {a \left (a^2-3 b^2\right ) (n p+2) \tan ^{n p+1}(e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (n p+1),\frac {1}{2} (n p+3),-\tan ^2(e+f x)\right )}{n p+1}+b \left (3 a^2-b^2\right ) \tan ^{n p+2}(e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (n p+2),\frac {1}{2} (n p+4),-\tan ^2(e+f x)\right )+\frac {3 a b^2 (n p+2) \tan ^{n p+1}(e+f x)}{n p+1}}{n p+2}+\frac {b^3 \tan ^{n p+2}(e+f x)}{n p+2}\right )}{f}\) |
((c*(d*Tan[e + f*x])^p)^n*((b^3*Tan[e + f*x]^(2 + n*p))/(2 + n*p) + ((3*a* b^2*(2 + n*p)*Tan[e + f*x]^(1 + n*p))/(1 + n*p) + (a*(a^2 - 3*b^2)*(2 + n* p)*Hypergeometric2F1[1, (1 + n*p)/2, (3 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]^(1 + n*p))/(1 + n*p) + b*(3*a^2 - b^2)*Hypergeometric2F1[1, (2 + n*p )/2, (4 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]^(2 + n*p))/(2 + n*p)))/(f* Tan[e + f*x]^(n*p))
3.14.24.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1)) Int[(e*x)^m*(a + b* x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 )*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && IGtQ[n, 1] && !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
Int[(u_.)*((c_.)*((d_)*((a_.) + (b_.)*(x_)))^(q_))^(p_), x_Symbol] :> Simp[ (c*(d*(a + b*x))^q)^p/(a + b*x)^(p*q) Int[u*(a + b*x)^(p*q), x], x] /; Fr eeQ[{a, b, c, d, q, p}, x] && !IntegerQ[q] && !IntegerQ[p]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Tan[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], x ]]
\[\int \left (c \left (d \tan \left (f x +e \right )\right )^{p}\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{3}d x\]
\[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n} \,d x } \]
integral((b^3*tan(f*x + e)^3 + 3*a*b^2*tan(f*x + e)^2 + 3*a^2*b*tan(f*x + e) + a^3)*((d*tan(f*x + e))^p*c)^n, x)
\[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\int \left (c \left (d \tan {\left (e + f x \right )}\right )^{p}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \]
\[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n} \,d x } \]
\[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n} \,d x } \]
Timed out. \[ \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^3 \, dx=\int {\left (c\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \]