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

Optimal. Leaf size=102 \[ \frac{\sec ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{p+1}}{2 d (a+b)}-\frac{(a+b p+b) \left (a+b \sin ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^2(c+d x)+a}{a+b}\right )}{2 d (p+1) (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.0936865, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3194, 78, 68} \[ \frac{\sec ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{p+1}}{2 d (a+b)}-\frac{(a+b p+b) \left (a+b \sin ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^2(c+d x)+a}{a+b}\right )}{2 d (p+1) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(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*Sin[c + d*x]^2)^(1 +
p))/(2*(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 3194

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]

Rule 78

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] || LtQ
[p, n]))))

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 ^2(c+d x)\right )^p \tan ^3(c+d x) \, dx &=\frac{\operatorname{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) \operatorname{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) \, _2F_1\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]  time = 0.235893, size = 83, normalized size = 0.81 \[ \frac{\left (a+b \sin ^2(c+d x)\right )^{p+1} \left ((p+1) (a+b) \sec ^2(c+d x)-(a+b p+b) \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^2(c+d x)+a}{a+b}\right )\right )}{2 d (p+1) (a+b)^2} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.889, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ \left ( \sin \left ( dx+c \right ) \right ) ^{2}b \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+sin(d*x+c)^2*b)^p*tan(d*x+c)^3,x)

[Out]

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

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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 )^{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)^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)

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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)**2)**p*tan(d*x+c)**3,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 )^{2} + 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)^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)