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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3308, 849, 858, 252, 251, 771, 441, 440, 455, 70} \[ \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx=-\frac {(a+2 b p+b) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}-\frac {(a+2 b p+b) \left (a+b \sin ^4(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)^2}+\frac {b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}+\frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{p+1}}{2 d (a+b)} \]

[In]

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

[Out]

-1/4*((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))/((a + b)^2*d*(1 + p)) + (Sec[c + d*x]^2*(a + b*Sin[c + d*x]^4)^(1 + p))/(2*(a + b)*d) - ((a + b + 2
*b*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)/(2*(a + b)*d*(1 + (b*Sin[c + d*x]^4)/a)^p) + (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)/(2*(a + b)*d*(1 + (b*Sin[c + d*x]^4)/a)^p)

Rule 70

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 251

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

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 440

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

Rule 441

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

Rule 455

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

Rule 771

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - e*(x/(d^2 - e^2*x^2)))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 849

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*
(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(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] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 3308

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x]^2, x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x \left (a+b x^2\right )^p}{(1-x)^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d} \\ & = \frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{2 (a+b) d}-\frac {\text {Subst}\left (\int \frac {(a+b (1+2 p) x) \left (a+b x^2\right )^p}{1-x} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d} \\ & = \frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{2 (a+b) d}+\frac {(b (1+2 p)) \text {Subst}\left (\int \left (a+b x^2\right )^p \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d}-\frac {(a+b+2 b p) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{1-x} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d} \\ & = \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) \text {Subst}\left (\int \left (\frac {\left (a+b x^2\right )^p}{1-x^2}-\frac {x \left (a+b x^2\right )^p}{-1+x^2}\right ) \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d}+\frac {\left (b (1+2 p) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \left (1+\frac {b x^2}{a}\right )^p \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d} \\ & = \frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{1+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}-\frac {(a+b+2 b p) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{1-x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d}+\frac {(a+b+2 b p) \text {Subst}\left (\int \frac {x \left (a+b x^2\right )^p}{-1+x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d} \\ & = \frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{1+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}+\frac {(a+b+2 b p) \text {Subst}\left (\int \frac {(a+b x)^p}{-1+x} \, dx,x,\sin ^4(c+d x)\right )}{4 (a+b) d}-\frac {\left ((a+b+2 b p) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{1-x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d} \\ & = -\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} \\ \end{align*}

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

[In]

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

[Out]

((-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 + Sqrt[-(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*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[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 + Sqr
t[-(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[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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