\(\int (d \sin (e+f x))^m (b \tan ^2(e+f x))^p \, dx\) [152]

Optimal result

Integrand size = 23, antiderivative size = 92 \[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\frac {\cos ^2(e+f x)^{\frac {1}{2}+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 p),\frac {1}{2} (1+m+2 p),\frac {1}{2} (3+m+2 p),\sin ^2(e+f x)\right ) (d \sin (e+f x))^m \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{f (1+m+2 p)} \] Output:

(cos(f*x+e)^2)^(1/2+p)*hypergeom([1/2+p, 1/2+1/2*m+p],[3/2+1/2*m+p],sin(f* 
x+e)^2)*(d*sin(f*x+e))^m*tan(f*x+e)*(b*tan(f*x+e)^2)^p/f/(1+m+2*p)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 1.97 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.17 \[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\frac {(3+m+2 p) \operatorname {AppellF1}\left (\frac {1}{2}+\frac {m}{2}+p,2 p,1+m,\frac {3}{2}+\frac {m}{2}+p,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x) (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p}{f (1+m+2 p) \left ((3+m+2 p) \operatorname {AppellF1}\left (\frac {1}{2}+\frac {m}{2}+p,2 p,1+m,\frac {3}{2}+\frac {m}{2}+p,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((1+m) \operatorname {AppellF1}\left (\frac {3}{2}+\frac {m}{2}+p,2 p,2+m,\frac {5}{2}+\frac {m}{2}+p,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 p \operatorname {AppellF1}\left (\frac {3}{2}+\frac {m}{2}+p,1+2 p,1+m,\frac {5}{2}+\frac {m}{2}+p,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(d*Sin[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]
 

Output:

((3 + m + 2*p)*AppellF1[1/2 + m/2 + p, 2*p, 1 + m, 3/2 + m/2 + p, Tan[(e + 
 f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[e + f*x]*(d*Sin[e + f*x])^m*(b*Tan[e 
+ f*x]^2)^p)/(f*(1 + m + 2*p)*((3 + m + 2*p)*AppellF1[1/2 + m/2 + p, 2*p, 
1 + m, 3/2 + m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((1 + m 
)*AppellF1[3/2 + m/2 + p, 2*p, 2 + m, 5/2 + m/2 + p, Tan[(e + f*x)/2]^2, - 
Tan[(e + f*x)/2]^2] - 2*p*AppellF1[3/2 + m/2 + p, 1 + 2*p, 1 + m, 5/2 + m/ 
2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4141, 3042, 3082, 3042, 3057}

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.

\(\displaystyle \int \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (b \tan (e+f x)^2\right )^p (d \sin (e+f x))^mdx\)

\(\Big \downarrow \) 4141

\(\displaystyle \tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p \int (d \sin (e+f x))^m \tan ^{2 p}(e+f x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p \int (d \sin (e+f x))^m \tan (e+f x)^{2 p}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle d \sin (e+f x) \cos ^{2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^{-2 p-1} \int \cos ^{-2 p}(e+f x) (d \sin (e+f x))^{m+2 p}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle d \sin (e+f x) \cos ^{2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^{-2 p-1} \int \cos (e+f x)^{-2 p} (d \sin (e+f x))^{m+2 p}dx\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {\tan (e+f x) \cos ^2(e+f x)^{p+\frac {1}{2}} \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 p+1),\frac {1}{2} (m+2 p+1),\frac {1}{2} (m+2 p+3),\sin ^2(e+f x)\right )}{f (m+2 p+1)}\)

Input:

Int[(d*Sin[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]
 

Output:

((Cos[e + f*x]^2)^(1/2 + p)*Hypergeometric2F1[(1 + 2*p)/2, (1 + m + 2*p)/2 
, (3 + m + 2*p)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^m*Tan[e + f*x]*(b*Tan[ 
e + f*x]^2)^p)/(f*(1 + m + 2*p))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac 
Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr 
acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ 
e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
 

rule 3082

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[a*Cos[e + f*x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b* 
(a*Sin[e + f*x])^(n + 1)))   Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x 
], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]
 

rule 4141

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)
 

Output:

int((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\int { \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*tan(f*x + e)^2)^p*(d*sin(f*x + e))^m, x)
 

Sympy [F]

\[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\int \left (b \tan ^{2}{\left (e + f x \right )}\right )^{p} \left (d \sin {\left (e + f x \right )}\right )^{m}\, dx \] Input:

integrate((d*sin(f*x+e))**m*(b*tan(f*x+e)**2)**p,x)
 

Output:

Integral((b*tan(e + f*x)**2)**p*(d*sin(e + f*x))**m, x)
 

Maxima [F]

\[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\int { \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e)^2)^p*(d*sin(f*x + e))^m, x)
 

Giac [F]

\[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\int { \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e)^2)^p*(d*sin(f*x + e))^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^p \,d x \] Input:

int((d*sin(e + f*x))^m*(b*tan(e + f*x)^2)^p,x)
 

Output:

int((d*sin(e + f*x))^m*(b*tan(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int (d \sin (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=d^{m} b^{p} \left (\int \tan \left (f x +e \right )^{2 p} \sin \left (f x +e \right )^{m}d x \right ) \] Input:

int((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)
 

Output:

d**m*b**p*int(tan(e + f*x)**(2*p)*sin(e + f*x)**m,x)