3.565 \(\int (a+b \sin ^4(c+d x))^p \tan ^3(c+d x) \, dx\)
Optimal. Leaf size=279 \[ -\frac{(a+2 b p+b) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};\sin ^4(c+d x),-\frac{b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}+\frac{b (2 p+1) \sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sin ^4(c+d x)}{a}\right )}{2 d (a+b)}-\frac{(a+2 b p+b) \left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)^2}+\frac{\sec ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^{p+1}}{2 d (a+b)} \]
[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)
________________________________________________________________________________________
Rubi [A] time = 0.293776, antiderivative size = 279, normalized size of antiderivative = 1.,
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 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]
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 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 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 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 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 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 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 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]
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 = 17.2688, size = 2007, normalized size = 7.19 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Sin[c + d*x]^4)^p*Tan[c + d*x]^3,x]
[Out]
-(((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)])] + 2*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*Tan[c + d*x]^3*
(Cos[c + d*x]^4*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))^p)/(4*d*p*(-1 + 2*p)*((-a + Sqrt[-(a*b)] -
(a + b)*Tan[c + d*x]^2)/(b + Sqrt[-(a*b)]))^p*((a + Sqrt[-(a*b)] + (a + b)*Tan[c + d*x]^2)/(-b + Sqrt[-(a*b)])
)^p*(((a + b)*Sec[c + d*x]^2*((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)])] + 2*p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, -(((a + b)*Se
c[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])]*Sec[c + d*x]^2)*Tan[c + d*x]*
((a + Sqrt[-(a*b)] + (a + b)*Tan[c + d*x]^2)/(-b + Sqrt[-(a*b)]))^(-1 - p)*(Cos[c + d*x]^4*(a + 2*a*Tan[c + d*
x]^2 + (a + b)*Tan[c + d*x]^4))^p)/(2*(-b + Sqrt[-(a*b)])*(-1 + 2*p)*((-a + Sqrt[-(a*b)] - (a + b)*Tan[c + d*x
]^2)/(b + Sqrt[-(a*b)]))^p) - ((a + b)*Sec[c + d*x]^2*((1 - 2*p)*AppellF1[-2*p, -p, -p, 1 - 2*p, -(((a + b)*Se
c[c + d*x]^2)/(-b + Sqrt[-(a*b)])), ((a + b)*Sec[c + d*x]^2)/(b + Sqrt[-(a*b)])] + 2*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)*Tan[c + d*x]*((-a + Sqrt[-(a*b)] - (a + b)*Tan[c + d*x]^2)/(b + Sqrt[-(a*b)]))^(-1 - p)*(Cos[c + d
*x]^4*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))^p)/(2*(b + Sqrt[-(a*b)])*(-1 + 2*p)*((a + Sqrt[-(a*b)
] + (a + b)*Tan[c + d*x]^2)/(-b + Sqrt[-(a*b)]))^p) - ((Cos[c + d*x]^4*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c
+ d*x]^4))^p*(4*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*Tan[c + d*x] + (1 - 2*p)*((-4*(a + b)*p^2*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)])]*Sec[c + d*x]^2*Tan[c + d*x])/((-b + Sqrt[-(a*b)])*(1 - 2*p)) + (4*(a + b)*p^2*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)])]*Sec
[c + d*x]^2*Tan[c + d*x])/((b + Sqrt[-(a*b)])*(1 - 2*p))) + 2*p*Sec[c + d*x]^2*((2*(a + b)*(1 - 2*p)*p*AppellF
1[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)])]*Sec[c + d*x]^2*Tan[c + d*x])/((-b + Sqrt[-(a*b)])*(2 - 2*p)) - (2*(a + b)*(1 - 2*p)*p*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 + Sq
rt[-(a*b)])]*Sec[c + d*x]^2*Tan[c + d*x])/((b + Sqrt[-(a*b)])*(2 - 2*p)))))/(4*p*(-1 + 2*p)*((-a + Sqrt[-(a*b)
] - (a + b)*Tan[c + d*x]^2)/(b + Sqrt[-(a*b)]))^p*((a + Sqrt[-(a*b)] + (a + b)*Tan[c + d*x]^2)/(-b + Sqrt[-(a*
b)]))^p) - (((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)])] + 2*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)*(Cos[c + d*x]^4*(a + 2*a*Tan[c
+ d*x]^2 + (a + b)*Tan[c + d*x]^4))^(-1 + p)*(Cos[c + d*x]^4*(4*a*Sec[c + d*x]^2*Tan[c + d*x] + 4*(a + b)*Sec
[c + d*x]^2*Tan[c + d*x]^3) - 4*Cos[c + d*x]^3*Sin[c + d*x]*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))
)/(4*(-1 + 2*p)*((-a + Sqrt[-(a*b)] - (a + b)*Tan[c + d*x]^2)/(b + Sqrt[-(a*b)]))^p*((a + Sqrt[-(a*b)] + (a +
b)*Tan[c + d*x]^2)/(-b + Sqrt[-(a*b)]))^p)))
________________________________________________________________________________________
Maple [F] time = 1.094, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4} \right ) ^{p} \left ( \tan \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}
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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}
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)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
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)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}
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)