\(\int (a+b \sin ^2(c+d x))^p \tan ^3(c+d x) \, dx\) [544]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=-\frac {(a+b+b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sin ^2(c+d x)}{a+b}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b)^2 d (1+p)}+\frac {\sec ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3273, 79, 70} \[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\frac {\sec ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{p+1}}{2 d (a+b)}-\frac {(a+b p+b) \left (a+b \sin ^2(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^2(c+d x)+a}{a+b}\right )}{2 d (p+1) (a+b)^2} \]

[In]

Int[(a + b*Sin[c + d*x]^2)^p*Tan[c + d*x]^3,x]

[Out]

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

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 3273

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x (a+b x)^p}{(1-x)^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d} \\ & = \frac {\sec ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b) d}-\frac {(a+b+b p) \text {Subst}\left (\int \frac {(a+b x)^p}{1-x} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d} \\ & = -\frac {(a+b+b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sin ^2(c+d x)}{a+b}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b)^2 d (1+p)}+\frac {\sec ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\frac {\left (-\left ((a+b+b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sin ^2(c+d x)}{a+b}\right )\right )+(a+b) (1+p) \sec ^2(c+d x)\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b)^2 d (1+p)} \]

[In]

Integrate[(a + b*Sin[c + d*x]^2)^p*Tan[c + d*x]^3,x]

[Out]

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

Maple [F]

\[\int {\left (a +\left (\sin ^{2}\left (d x +c \right )\right ) b \right )}^{p} \left (\tan ^{3}\left (d x +c \right )\right )d x\]

[In]

int((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^3,x)

[Out]

int((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^3,x)

Fricas [F]

\[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*sin(d*x+c)**2)**p*tan(d*x+c)**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*tan(d*x + c)^3, x)

Giac [F]

\[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*tan(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^2+a\right )}^p \,d x \]

[In]

int(tan(c + d*x)^3*(a + b*sin(c + d*x)^2)^p,x)

[Out]

int(tan(c + d*x)^3*(a + b*sin(c + d*x)^2)^p, x)