3.14.25 \(\int (c (d \tan (e+f x))^p)^n (a+b \tan (e+f x))^2 \, dx\) [1325]

Optimal. Leaf size=171 \[ \frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {\left (a^2-b^2\right ) \, _2F_1\left (1,\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {2 a b \, _2F_1\left (1,\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);-\tan ^2(e+f x)\right ) \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (2+n p)} \]

[Out]

b^2*tan(f*x+e)*(c*(d*tan(f*x+e))^p)^n/f/(n*p+1)+(a^2-b^2)*hypergeom([1, 1/2*n*p+1/2],[1/2*n*p+3/2],-tan(f*x+e)
^2)*tan(f*x+e)*(c*(d*tan(f*x+e))^p)^n/f/(n*p+1)+2*a*b*hypergeom([1, 1/2*n*p+1],[1/2*n*p+2],-tan(f*x+e)^2)*tan(
f*x+e)^2*(c*(d*tan(f*x+e))^p)^n/f/(n*p+2)

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Rubi [A]
time = 0.17, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1970, 1816, 822, 371} \begin {gather*} \frac {\left (a^2-b^2\right ) \tan (e+f x) \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right ) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)}+\frac {2 a b \tan ^2(e+f x) \, _2F_1\left (1,\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);-\tan ^2(e+f x)\right ) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+2)}+\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*(d*Tan[e + f*x])^p)^n*(a + b*Tan[e + f*x])^2,x]

[Out]

(b^2*Tan[e + f*x]*(c*(d*Tan[e + f*x])^p)^n)/(f*(1 + n*p)) + ((a^2 - b^2)*Hypergeometric2F1[1, (1 + n*p)/2, (3
+ n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(c*(d*Tan[e + f*x])^p)^n)/(f*(1 + n*p)) + (2*a*b*Hypergeometric2F1[1,
(2 + n*p)/2, (4 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]^2*(c*(d*Tan[e + f*x])^p)^n)/(f*(2 + n*p))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1970

Int[(u_.)*((c_.)*((d_)*((a_.) + (b_.)*(x_)))^(q_))^(p_), x_Symbol] :> Dist[(c*(d*(a + b*x))^q)^p/(a + b*x)^(p*
q), Int[u*(a + b*x)^(p*q), x], x] /; FreeQ[{a, b, c, d, q, p}, x] &&  !IntegerQ[q] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \left (c (d \tan (e+f x))^p\right )^n (a+b \tan (e+f x))^2 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (c (d x)^p\right )^n (a+b x)^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p} (a+b x)^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \left (b^2 (d x)^{n p}+\frac {(d x)^{n p} \left (a^2-b^2+2 a b x\right )}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p} \left (a^2-b^2+2 a b x\right )}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {\left (\left (a^2-b^2\right ) (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (2 a b (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{1+n p}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{d f}\\ &=\frac {b^2 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {\left (a^2-b^2\right ) \, _2F_1\left (1,\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac {2 a b \, _2F_1\left (1,\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);-\tan ^2(e+f x)\right ) \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (2+n p)}\\ \end {align*}

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Mathematica [A]
time = 0.59, size = 136, normalized size = 0.80 \begin {gather*} \frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n \left (\left (a^2-b^2\right ) (2+n p) \, _2F_1\left (1,\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);-\tan ^2(e+f x)\right )+b \left (b (2+n p)+2 a (1+n p) \, _2F_1\left (1,1+\frac {n p}{2};2+\frac {n p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{f (1+n p) (2+n p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*(d*Tan[e + f*x])^p)^n*(a + b*Tan[e + f*x])^2,x]

[Out]

(Tan[e + f*x]*(c*(d*Tan[e + f*x])^p)^n*((a^2 - b^2)*(2 + n*p)*Hypergeometric2F1[1, (1 + n*p)/2, (3 + n*p)/2, -
Tan[e + f*x]^2] + b*(b*(2 + n*p) + 2*a*(1 + n*p)*Hypergeometric2F1[1, 1 + (n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^
2]*Tan[e + f*x])))/(f*(1 + n*p)*(2 + n*p))

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \left (c \left (d \tan \left (f x +e \right )\right )^{p}\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*tan(f*x+e))^p)^n*(a+b*tan(f*x+e))^2,x)

[Out]

int((c*(d*tan(f*x+e))^p)^n*(a+b*tan(f*x+e))^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*tan(f*x+e))^p)^n*(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^2*((d*tan(f*x + e))^p*c)^n, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*tan(f*x+e))^p)^n*(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*((d*tan(f*x + e))^p*c)^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c \left (d \tan {\left (e + f x \right )}\right )^{p}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*tan(f*x+e))**p)**n*(a+b*tan(f*x+e))**2,x)

[Out]

Integral((c*(d*tan(e + f*x))**p)**n*(a + b*tan(e + f*x))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*tan(f*x+e))^p)^n*(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^2*((d*tan(f*x + e))^p*c)^n, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*tan(e + f*x))^p)^n*(a + b*tan(e + f*x))^2,x)

[Out]

int((c*(d*tan(e + f*x))^p)^n*(a + b*tan(e + f*x))^2, x)

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