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

Optimal. Leaf size=141 \[ \frac{\sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};\sin ^4(c+d x),-\frac{b \sin ^4(c+d x)}{a}\right )}{2 d}+\frac{\left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.119269, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3229, 757, 430, 429, 444, 68} \[ \frac{\sin ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};\sin ^4(c+d x),-\frac{b \sin ^4(c+d x)}{a}\right )}{2 d}+\frac{\left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^4(c+d x)+a}{a+b}\right )}{4 d (p+1) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)
*d*(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)/(2*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 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 (c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{1-x} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac{\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 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{1-x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}-\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{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{-1+x} \, dx,x,\sin ^4(c+d x)\right )}{4 d}+\frac{\left (\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 d}\\ &=\frac{\, _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) d (1+p)}+\frac{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 d}\\ \end{align*}

Mathematica [B]  time = 8.08056, size = 466, normalized size = 3.3 \[ -\frac{(2 p-1) \left (\sqrt{-a b}-b\right ) \left (\sqrt{-a b}+b\right ) \sin (c+d x) \cos (c+d x) \left (-(a+b) \tan ^2(c+d x)+\sqrt{-a b}-a\right ) \left ((a+b) \tan ^2(c+d x)+\sqrt{-a b}+a\right ) \left (a+b \sin ^4(c+d x)\right )^p F_1\left (-2 p;-p,-p;1-2 p;-\frac{(a+b) \sec ^2(c+d x)}{\sqrt{-a b}-b},\frac{(a+b) \sec ^2(c+d x)}{b+\sqrt{-a b}}\right )}{2 d p (a+b)^2 \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right ) \left (b (2 p-1) \sin (2 (c+d x)) F_1\left (-2 p;-p,-p;1-2 p;-\frac{(a+b) \sec ^2(c+d x)}{\sqrt{-a b}-b},\frac{(a+b) \sec ^2(c+d x)}{b+\sqrt{-a b}}\right )+2 p \tan (c+d x) \left (\left (\sqrt{-a b}+b\right ) F_1\left (1-2 p;1-p,-p;2-2 p;-\frac{(a+b) \sec ^2(c+d x)}{\sqrt{-a b}-b},\frac{(a+b) \sec ^2(c+d x)}{b+\sqrt{-a b}}\right )+\left (b-\sqrt{-a b}\right ) F_1\left (1-2 p;-p,1-p;2-2 p;-\frac{(a+b) \sec ^2(c+d x)}{\sqrt{-a b}-b},\frac{(a+b) \sec ^2(c+d x)}{b+\sqrt{-a b}}\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-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]*Sin[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*(a + b)^
2*d*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 - Sq
rt[-(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)*Se
c[c + d*x]^2)/(b + Sqrt[-(a*b)])])*Tan[c + d*x])*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))

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Maple [F]  time = 1.684, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4} \right ) ^{p}\tan \left ( dx+c \right ) \, 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),x)

[Out]

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

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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 )\,{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),x, algorithm="maxima")

[Out]

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

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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 ), 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),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), x)

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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),x)

[Out]

Timed out

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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 )\,{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),x, algorithm="giac")

[Out]

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