3.6.65 \(\int (a+b \sin ^4(c+d x))^p \tan ^3(c+d x) \, dx\) [565]

3.6.65.1 Optimal result

Integrand size = 23, antiderivative size = 279 \[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=-\frac {(a+b+2 b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sin ^4(c+d x)}{a+b}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 (a+b)^2 d (1+p)}+\frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{2 (a+b) d}-\frac {(a+b+2 b p) \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right ) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}}{2 (a+b) d}+\frac {b (1+2 p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right ) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}}{2 (a+b) d} \]

output

-1/4*(2*b*p+a+b)*hypergeom([1, p+1],[2+p],(a+b*sin(d*x+c)^4)/(a+b))*(a+b*s 
in(d*x+c)^4)^(p+1)/(a+b)^2/d/(p+1)+1/2*sec(d*x+c)^2*(a+b*sin(d*x+c)^4)^(p+ 
1)/(a+b)/d-1/2*(2*b*p+a+b)*AppellF1(1/2,1,-p,3/2,sin(d*x+c)^4,-b*sin(d*x+c 
)^4/a)*sin(d*x+c)^2*(a+b*sin(d*x+c)^4)^p/(a+b)/d/((1+b*sin(d*x+c)^4/a)^p)+ 
1/2*b*(1+2*p)*hypergeom([1/2, -p],[3/2],-b*sin(d*x+c)^4/a)*sin(d*x+c)^2*(a 
+b*sin(d*x+c)^4)^p/(a+b)/d/((1+b*sin(d*x+c)^4/a)^p)
 

3.6.65.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(810\) vs. \(2(279)=558\).

Time = 14.11 (sec) , antiderivative size = 810, normalized size of antiderivative = 2.90 \[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=\frac {\left (-b+\sqrt {-a b}\right ) \left (b+\sqrt {-a b}\right ) \cos (c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (-a+\sqrt {-a b}-(a+b) \tan ^2(c+d x)\right ) \left (a+\sqrt {-a b}+(a+b) \tan ^2(c+d x)\right ) \left (-\frac {2 (-1+p) \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \sec ^3(c+d x)}{2 b (-1+p) \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )+p \left (\left (b+\sqrt {-a b}\right ) \operatorname {AppellF1}\left (2-2 p,1-p,-p,3-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )+\left (b-\sqrt {-a b}\right ) \operatorname {AppellF1}\left (2-2 p,-p,1-p,3-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )\right ) \sec ^2(c+d x)}+\frac {(1-2 p)^2 \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \sin (c+d x)}{p \left (b (-1+2 p) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \sin (2 (c+d x))+2 p \left (\left (b+\sqrt {-a b}\right ) \operatorname {AppellF1}\left (1-2 p,1-p,-p,2-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )+\left (b-\sqrt {-a b}\right ) \operatorname {AppellF1}\left (1-2 p,-p,1-p,2-2 p,-\frac {(a+b) \sec ^2(c+d x)}{-b+\sqrt {-a b}},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )\right ) \tan (c+d x)\right )}\right )}{2 (a+b)^2 d (-1+2 p) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )} \]

input

Integrate[(a + b*Sin[c + d*x]^4)^p*Tan[c + d*x]^3,x]
 

output

((-b + Sqrt[-(a*b)])*(b + Sqrt[-(a*b)])*Cos[c + d*x]*(a + b*Sin[c + d*x]^4 
)^p*(-a + Sqrt[-(a*b)] - (a + b)*Tan[c + d*x]^2)*(a + Sqrt[-(a*b)] + (a + 
b)*Tan[c + d*x]^2)*((-2*(-1 + p)*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, -(((a 
+ b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + S 
qrt[-(a*b)])]*Sec[c + d*x]^3)/(2*b*(-1 + p)*AppellF1[1 - 2*p, -p, -p, 2 - 
2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x 
]^2)/(b + Sqrt[-(a*b)])] + p*((b + Sqrt[-(a*b)])*AppellF1[2 - 2*p, 1 - p, 
-p, 3 - 2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec 
[c + d*x]^2)/(b + Sqrt[-(a*b)])] + (b - Sqrt[-(a*b)])*AppellF1[2 - 2*p, -p 
, 1 - p, 3 - 2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b 
)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])])*Sec[c + d*x]^2) + ((1 - 2*p)^2*Appe 
llF1[-2*p, -p, -p, 1 - 2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)]) 
), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])]*Sin[c + d*x])/(p*(b*(-1 + 
2*p)*AppellF1[-2*p, -p, -p, 1 - 2*p, -(((a + b)*Sec[c + d*x]^2)/(-b + Sqrt 
[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])]*Sin[2*(c + d*x)] 
+ 2*p*((b + Sqrt[-(a*b)])*AppellF1[1 - 2*p, 1 - p, -p, 2 - 2*p, -(((a + b) 
*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[ 
-(a*b)])] + (b - Sqrt[-(a*b)])*AppellF1[1 - 2*p, -p, 1 - p, 2 - 2*p, -(((a 
 + b)*Sec[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + 
Sqrt[-(a*b)])])*Tan[c + d*x]))))/(2*(a + b)^2*d*(-1 + 2*p)*(a + 2*a*Tan...
 

3.6.65.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3708, 594, 719, 238, 237, 504, 334, 333, 353, 78}

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.

\(\displaystyle \int \tan ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^3 \left (a+b \sin (c+d x)^4\right )^pdx\)

\(\Big \downarrow \) 3708

\(\displaystyle \frac {\int \frac {\sin ^2(c+d x) \left (b \sin ^4(c+d x)+a\right )^p}{\left (1-\sin ^2(c+d x)\right )^2}d\sin ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 594

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {\int \frac {\left (b (2 p+1) \sin ^2(c+d x)+a\right ) \left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^2(c+d x)}d\sin ^2(c+d x)}{a+b}}{2 d}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \int \frac {\left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^2(c+d x)}d\sin ^2(c+d x)-b (2 p+1) \int \left (b \sin ^4(c+d x)+a\right )^pd\sin ^2(c+d x)}{a+b}}{2 d}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \int \frac {\left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^2(c+d x)}d\sin ^2(c+d x)-b (2 p+1) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \int \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^pd\sin ^2(c+d x)}{a+b}}{2 d}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \int \frac {\left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^2(c+d x)}d\sin ^2(c+d x)-b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{a+b}}{2 d}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \left (\int \frac {\left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^4(c+d x)}d\sin ^2(c+d x)+\int \frac {\sin ^2(c+d x) \left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^4(c+d x)}d\sin ^2(c+d x)\right )-b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{a+b}}{2 d}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \left (\left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \int \frac {\left (\frac {b \sin ^4(c+d x)}{a}+1\right )^p}{1-\sin ^4(c+d x)}d\sin ^2(c+d x)+\int \frac {\sin ^2(c+d x) \left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^4(c+d x)}d\sin ^2(c+d x)\right )-b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{a+b}}{2 d}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \left (\int \frac {\sin ^2(c+d x) \left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^4(c+d x)}d\sin ^2(c+d x)+\sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )\right )-b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{a+b}}{2 d}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \left (\frac {1}{2} \int \frac {\left (b \sin ^4(c+d x)+a\right )^p}{1-\sin ^4(c+d x)}d\sin ^4(c+d x)+\sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )\right )-b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{a+b}}{2 d}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {(a+2 b p+b) \left (\sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )+\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^4(c+d x)+a}{a+b}\right )}{2 (p+1) (a+b)}\right )-b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{a+b}}{2 d}\)

input

Int[(a + b*Sin[c + d*x]^4)^p*Tan[c + d*x]^3,x]
 

output

((a + b*Sin[c + d*x]^4)^(1 + p)/((a + b)*(1 - Sin[c + d*x]^2)) - (-((b*(1 
+ 2*p)*Hypergeometric2F1[1/2, -p, 3/2, -((b*Sin[c + d*x]^4)/a)]*Sin[c + d* 
x]^2*(a + b*Sin[c + d*x]^4)^p)/(1 + (b*Sin[c + d*x]^4)/a)^p) + (a + b + 2* 
b*p)*((Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sin[c + d*x]^4)/(a + b)]* 
(a + b*Sin[c + d*x]^4)^(1 + p))/(2*(a + b)*(1 + p)) + (AppellF1[1/2, 1, -p 
, 3/2, Sin[c + d*x]^4, -((b*Sin[c + d*x]^4)/a)]*Sin[c + d*x]^2*(a + b*Sin[ 
c + d*x]^4)^p)/(1 + (b*Sin[c + d*x]^4)/a)^p))/(a + b))/(2*d)
 

3.6.65.3.1 Defintions of rubi rules used

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 3042

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

rule 3708

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(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^(n/2)*x^(n/2))^p/(1 
- ff*x)^((m + 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, 
p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2]
 

3.6.65.4 Maple [F]

\[\int {\left (a +b \left (\sin ^{4}\left (d x +c \right )\right )\right )}^{p} \left (\tan ^{3}\left (d x +c \right )\right )d x\]

input

int((a+b*sin(d*x+c)^4)^p*tan(d*x+c)^3,x)
 

output

int((a+b*sin(d*x+c)^4)^p*tan(d*x+c)^3,x)
 

3.6.65.5 Fricas [F]

\[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int { {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3} \,d x } \]

input

integrate((a+b*sin(d*x+c)^4)^p*tan(d*x+c)^3,x, algorithm="fricas")
 

output

integral((b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)^p*tan(d*x + c)^3, 
 x)
 

3.6.65.6 Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=\text {Timed out} \]

input

integrate((a+b*sin(d*x+c)**4)**p*tan(d*x+c)**3,x)
 

output

Timed out
 

3.6.65.7 Maxima [F]

\[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int { {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3} \,d x } \]

input

integrate((a+b*sin(d*x+c)^4)^p*tan(d*x+c)^3,x, algorithm="maxima")
 

output

integrate((b*sin(d*x + c)^4 + a)^p*tan(d*x + c)^3, x)
 

3.6.65.8 Giac [F]

\[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int { {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3} \,d x } \]

input

integrate((a+b*sin(d*x+c)^4)^p*tan(d*x+c)^3,x, algorithm="giac")
 

output

integrate((b*sin(d*x + c)^4 + a)^p*tan(d*x + c)^3, x)
 

3.6.65.9 Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^4+a\right )}^p \,d x \]

input

int(tan(c + d*x)^3*(a + b*sin(c + d*x)^4)^p,x)
 

output

int(tan(c + d*x)^3*(a + b*sin(c + d*x)^4)^p, x)