3.228 \(\int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=215 \[ \frac{2 (2 B n+i A (2 n+1)) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},-i \tan (c+d x)\right )}{d (2 n+1)}-\frac{2 (B+i A) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d}+\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)} \]

[Out]

(2*B*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + 2*n)) - (2*(I*A + B)*AppellF1[1/2, 1 - n, 1, 3/2, (-
I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + I*Tan[c + d*x])^n) + (2*
(2*B*n + I*A*(1 + 2*n))*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[
c + d*x])^n)/(d*(1 + 2*n)*(1 + I*Tan[c + d*x])^n)

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Rubi [A]  time = 0.494958, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3597, 3601, 3564, 130, 430, 429, 3599, 66, 64} \[ -\frac{2 (B+i A) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d}+\frac{2 (2 B n+i A (2 n+1)) \sqrt{\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right )}{d (2 n+1)}+\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

(2*B*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + 2*n)) - (2*(I*A + B)*AppellF1[1/2, 1 - n, 1, 3/2, (-
I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + I*Tan[c + d*x])^n) + (2*
(2*B*n + I*A*(1 + 2*n))*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[
c + d*x])^n)/(d*(1 + 2*n)*(1 + I*Tan[c + d*x])^n)

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rule 3564

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dis
t[(a*b)/f, Subst[Int[((a + x)^(m - 1)*(c + (d*x)/b)^n)/(b^2 + a*x), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b,
c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

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 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c^IntPart[n]*(c + d*x)^FracPart[n])/(1 + (d
*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0]))
||  !RationalQ[n])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac{2 \int \frac{(a+i a \tan (c+d x))^n \left (-\frac{a B}{2}+\frac{1}{2} a (A+2 A n-2 i B n) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{a (1+2 n)}\\ &=\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+(-i A-B) \int \frac{(a+i a \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx+\frac{(2 B n+i A (1+2 n)) \int \frac{(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx}{a (1+2 n)}\\ &=\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac{\left (a^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+n}}{\sqrt{-\frac{i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac{(a (2 B n+i A (1+2 n))) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-1+n}}{\sqrt{x}} \, dx,x,\tan (c+d x)\right )}{d (1+2 n)}\\ &=\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac{\left (2 a^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{\left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left ((2 B n+i A (1+2 n)) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(1+i x)^{-1+n}}{\sqrt{x}} \, dx,x,\tan (c+d x)\right )}{d (1+2 n)}\\ &=\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac{2 (2 B n+i A (1+2 n)) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac{\left (2 a^2 (i A+B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{\left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{2 B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}-\frac{2 (i A+B) F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}+\frac{2 (2 B n+i A (1+2 n)) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}\\ \end{align*}

Mathematica [F]  time = 20.2892, size = 0, normalized size = 0. \[ \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]), x]

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Maple [F]  time = 0.347, size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tan \left ( dx+c \right ) } \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sqrt{\tan \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*sqrt(tan(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(((A - I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*sqrt(
(-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))/(e^(2*I*d*x + 2*I*c) + 1), 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(tan(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**n*(A+B*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 \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sqrt{\tan \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*sqrt(tan(d*x + c)), x)