3.212 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=316 \[ \frac{2 a^2 \left (2 B \left (4 m^2+17 m+19\right )+i A \left (8 m^2+34 m+35\right )\right ) \sqrt{a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \text{Hypergeometric2F1}\left (\frac{1}{2},-m,\frac{3}{2},1+i \tan (c+d x)\right )}{d (2 m+3) (2 m+5)}+\frac{4 a^3 (A-i B) \sqrt{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt{a+i a \tan (c+d x)}}+\frac{2 a^2 (-A (2 m+5)+2 i B (m+4)) \sqrt{a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3) (2 m+5)}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2} \tan ^{m+1}(c+d x)}{d (2 m+5)} \]

[Out]

(4*a^3*(A - I*B)*AppellF1[1 + m, 1/2, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Ta
n[c + d*x]^(1 + m))/(d*(1 + m)*Sqrt[a + I*a*Tan[c + d*x]]) + (2*a^2*(2*B*(19 + 17*m + 4*m^2) + I*A*(35 + 34*m
+ 8*m^2))*Hypergeometric2F1[1/2, -m, 3/2, 1 + I*Tan[c + d*x]]*Tan[c + d*x]^m*Sqrt[a + I*a*Tan[c + d*x]])/(d*(3
+ 2*m)*(5 + 2*m)*((-I)*Tan[c + d*x])^m) + (2*a^2*((2*I)*B*(4 + m) - A*(5 + 2*m))*Tan[c + d*x]^(1 + m)*Sqrt[a
+ I*a*Tan[c + d*x]])/(d*(3 + 2*m)*(5 + 2*m)) + ((2*I)*a*B*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^(3/2))/(
d*(5 + 2*m))

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Rubi [A]  time = 0.986155, antiderivative size = 316, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3594, 3601, 3564, 135, 133, 3599, 67, 65} \[ \frac{4 a^3 (A-i B) \sqrt{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt{a+i a \tan (c+d x)}}+\frac{2 a^2 \left (2 B \left (4 m^2+17 m+19\right )+i A \left (8 m^2+34 m+35\right )\right ) \sqrt{a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};i \tan (c+d x)+1\right )}{d (2 m+3) (2 m+5)}+\frac{2 a^2 (-A (2 m+5)+2 i B (m+4)) \sqrt{a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3) (2 m+5)}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2} \tan ^{m+1}(c+d x)}{d (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(4*a^3*(A - I*B)*AppellF1[1 + m, 1/2, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Ta
n[c + d*x]^(1 + m))/(d*(1 + m)*Sqrt[a + I*a*Tan[c + d*x]]) + (2*a^2*(2*B*(19 + 17*m + 4*m^2) + I*A*(35 + 34*m
+ 8*m^2))*Hypergeometric2F1[1/2, -m, 3/2, 1 + I*Tan[c + d*x]]*Tan[c + d*x]^m*Sqrt[a + I*a*Tan[c + d*x]])/(d*(3
+ 2*m)*(5 + 2*m)*((-I)*Tan[c + d*x])^m) + (2*a^2*((2*I)*B*(4 + m) - A*(5 + 2*m))*Tan[c + d*x]^(1 + m)*Sqrt[a
+ I*a*Tan[c + d*x]])/(d*(3 + 2*m)*(5 + 2*m)) + ((2*I)*a*B*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^(3/2))/(
d*(5 + 2*m))

Rule 3594

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

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 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), 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*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 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 67

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

Rule 65

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

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac{2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac{2 \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} \left (-\frac{1}{2} a \left (2 i B (1+m)-2 A \left (\frac{5}{2}+m\right )\right )+\frac{1}{2} a (2 B (4+m)+i A (5+2 m)) \tan (c+d x)\right ) \, dx}{5+2 m}\\ &=\frac{2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt{a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac{2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac{4 \int \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 \left (2 i B \left (11+15 m+4 m^2\right )-A \left (25+30 m+8 m^2\right )\right )+\frac{1}{4} a^2 \left (2 B \left (19+17 m+4 m^2\right )+i A \left (35+34 m+8 m^2\right )\right ) \tan (c+d x)\right ) \, dx}{15+16 m+4 m^2}\\ &=\frac{2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt{a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac{2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\left (4 a^2 (A-i B)\right ) \int \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx+\frac{\left (a \left (2 i B \left (19+17 m+4 m^2\right )-A \left (35+34 m+8 m^2\right )\right )\right ) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx}{15+16 m+4 m^2}\\ &=\frac{2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt{a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac{2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac{\left (4 a^4 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a}\right )^m}{\sqrt{a+x} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac{\left (a^3 \left (2 i B \left (19+17 m+4 m^2\right )-A \left (35+34 m+8 m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^m}{\sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d \left (15+16 m+4 m^2\right )}\\ &=\frac{2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt{a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac{2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac{\left (a^3 \left (2 i B \left (19+17 m+4 m^2\right )-A \left (35+34 m+8 m^2\right )\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x)\right ) \operatorname{Subst}\left (\int \frac{(-i x)^m}{\sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d \left (15+16 m+4 m^2\right )}+\frac{\left (4 a^4 (i A+B) \sqrt{1+i \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a}\right )^m}{\sqrt{1+\frac{x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{4 a^3 (A-i B) F_1\left (1+m;\frac{1}{2},1;2+m;-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt{1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{d (1+m) \sqrt{a+i a \tan (c+d x)}}+\frac{2 a^2 \left (2 B \left (19+17 m+4 m^2\right )+i A \left (35+34 m+8 m^2\right )\right ) \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)}}{d \left (15+16 m+4 m^2\right )}+\frac{2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt{a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac{2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}\\ \end{align*}

Mathematica [F]  time = 6.68153, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]), x]

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Maple [F]  time = 0.51, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \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)^m*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(5/2)*(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 )}^{\frac{5}{2}} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{4 \, \sqrt{2}{\left ({\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (A + i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \left (\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^m*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(4*sqrt(2)*((A - I*B)*a^2*e^(6*I*d*x + 6*I*c) + (A + I*B)*a^2*e^(4*I*d*x + 4*I*c))*((-I*e^(2*I*d*x + 2
*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(e^(6*I*d*x + 6*I*c)
+ 3*e^(4*I*d*x + 4*I*c) + 3*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)**m*(a+I*a*tan(d*x+c))**(5/2)*(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 )}^{\frac{5}{2}} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^m, x)