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

Optimal. Leaf size=279 \[ -\frac {(a+2 b p+b) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}-\frac {(a+2 b p+b) \left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)^2}+\frac {b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}+\frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{p+1}}{2 d (a+b)} \]

[Out]

-1/4*(2*b*p+a+b)*hypergeom([1, 1+p],[2+p],(a+b*sin(d*x+c)^4)/(a+b))*(a+b*sin(d*x+c)^4)^(1+p)/(a+b)^2/d/(1+p)+1
/2*sec(d*x+c)^2*(a+b*sin(d*x+c)^4)^(1+p)/(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]  time = 0.29, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3229, 835, 844, 246, 245, 757, 430, 429, 444, 68} \[ -\frac {(a+2 b p+b) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\sin ^4(c+d x),-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}+\frac {b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}-\frac {(a+2 b p+b) \left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)^2}+\frac {\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{p+1}}{2 d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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))/(4*(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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 429

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 430

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 444

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 757

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 835

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 844

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 3229

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*} \int \left (a+b \sin ^4(c+d x)\right )^p \tan ^3(c+d x) \, dx &=\frac {\operatorname {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 {\operatorname {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)) \operatorname {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) \operatorname {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) \operatorname {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 ) \operatorname {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) \, _2F_1\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) \operatorname {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) \operatorname {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) \, _2F_1\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) \operatorname {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 ) \operatorname {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) \, _2F_1\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) F_1\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) \, _2F_1\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]  time = 18.56, size = 922, normalized size = 3.30 \[ -\frac {2 \left (\sqrt {-a b}-b\right ) \left (b+\sqrt {-a b}\right ) (2 p-1) F_1\left (-2 p;-p,-p;1-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \cos ^2(c+d x) \left (\left (a-\sqrt {-a b}\right ) \cot ^2(c+d x)+a+b\right ) \left (\left (a+\sqrt {-a b}\right ) \cot ^2(c+d x)+a+b\right ) \sin ^4(c+d x) \left (b \sin ^4(c+d x)+a\right )^p}{(a+b)^2 d p \left (b (2 p-1) F_1\left (-2 p;-p,-p;1-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \cos ^2(c+d x)+\left (b+\sqrt {-a b}\right ) p F_1\left (1-2 p;1-p,-p;2-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )-\left (\sqrt {-a b}-b\right ) p F_1\left (1-2 p;-p,1-p;2-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )\right ) (8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))}-\frac {\left (\sqrt {-a b}-b\right ) \left (b+\sqrt {-a b}\right ) (p-1) F_1\left (1-2 p;-p,-p;2-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right ) \sec ^2(c+d x) \left (-\left ((a+b) \tan ^2(c+d x)\right )-a+\sqrt {-a b}\right ) \left ((a+b) \tan ^2(c+d x)+a+\sqrt {-a b}\right ) \left (b \sin ^4(c+d x)+a\right )^p}{(a+b)^2 d (2 p-1) \left (p \left (\left (b+\sqrt {-a b}\right ) F_1\left (2-2 p;1-p,-p;3-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )+\left (b-\sqrt {-a b}\right ) F_1\left (2-2 p;-p,1-p;3-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )\right ) \sec ^2(c+d x)+2 b (p-1) F_1\left (1-2 p;-p,-p;2-2 p;-\frac {(a+b) \sec ^2(c+d x)}{\sqrt {-a b}-b},\frac {(a+b) \sec ^2(c+d x)}{b+\sqrt {-a b}}\right )\right ) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*(-b + Sqrt[-(a*b)])*(b + Sqrt[-(a*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)])]*Cos[c + d*x]^2*(a + b + (a - Sqrt[-(a*b)
])*Cot[c + d*x]^2)*(a + b + (a + Sqrt[-(a*b)])*Cot[c + d*x]^2)*Sin[c + d*x]^4*(a + b*Sin[c + d*x]^4)^p)/((a +
b)^2*d*p*((b + Sqrt[-(a*b)])*p*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)])*p*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)])] + 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)])]*Cos[c + d*x]^2)*(8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)])) - ((-b + Sqrt[-(a
*b)])*(b + Sqrt[-(a*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)])]*Sec[c + d*x]^2*(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))/((a + b)^2*d*(-1 + 2*p)*(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)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))

________________________________________________________________________________________

fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b\right )}^{p} \tan \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 2.62, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (\sin ^{4}\left (d x +c \right )\right )\right )^{p} \left (\tan ^{3}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^4+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________