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

Optimal. Leaf size=363 \[ \frac{(2 m+1) \left (B \left (-16 m^2+12 m+13\right )+i A \left (16 m^2-52 m+37\right )\right ) \sqrt{a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \text{Hypergeometric2F1}\left (\frac{1}{2},-m,\frac{3}{2},1+i \tan (c+d x)\right )}{60 a^3 d}+\frac{(A-i B) \sqrt{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{8 a^2 d (m+1) \sqrt{a+i a \tan (c+d x)}}-\frac{\left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((A + I*B)*Tan[c + d*x]^(1 + m))/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I*B*(1 - 4*m) + A*(11 - 4*m))*Tan[c +
d*x]^(1 + m))/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - ((I*B*(13 + 12*m - 16*m^2) - A*(37 - 52*m + 16*m^2))*Tan
[c + d*x]^(1 + m))/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((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]]*Tan[c + d*x]^(1 + m))/(8*a^2*d*(1 + m)*Sqrt[a + I*a*Tan[c +
d*x]]) + ((1 + 2*m)*(B*(13 + 12*m - 16*m^2) + I*A*(37 - 52*m + 16*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]])/(60*a^3*d*((-I)*Tan[c + d*x])^m)

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Rubi [A]  time = 1.36868, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3596, 3601, 3564, 135, 133, 3599, 67, 65} \[ \frac{(A-i B) \sqrt{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{8 a^2 d (m+1) \sqrt{a+i a \tan (c+d x)}}+\frac{(2 m+1) \left (B \left (-16 m^2+12 m+13\right )+i A \left (16 m^2-52 m+37\right )\right ) \sqrt{a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};i \tan (c+d x)+1\right )}{60 a^3 d}-\frac{\left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + I*B)*Tan[c + d*x]^(1 + m))/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I*B*(1 - 4*m) + A*(11 - 4*m))*Tan[c +
d*x]^(1 + m))/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - ((I*B*(13 + 12*m - 16*m^2) - A*(37 - 52*m + 16*m^2))*Tan
[c + d*x]^(1 + m))/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((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]]*Tan[c + d*x]^(1 + m))/(8*a^2*d*(1 + m)*Sqrt[a + I*a*Tan[c +
d*x]]) + ((1 + 2*m)*(B*(13 + 12*m - 16*m^2) + I*A*(37 - 52*m + 16*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]])/(60*a^3*d*((-I)*Tan[c + d*x])^m)

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 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 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 \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\tan ^m(c+d x) \left (a (A (4-m)-i B (1+m))-\frac{1}{2} a (i A-B) (3-2 m) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\tan ^m(c+d x) \left (-\frac{1}{2} a^2 \left (i B \left (7+3 m-4 m^2\right )-A \left (13-13 m+4 m^2\right )\right )+\frac{1}{4} a^2 (B (1-4 m)-i A (11-4 m)) (1-2 m) \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right ) \tan ^{1+m}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^3 \left (A \left (11+11 m-44 m^2+16 m^3\right )+i B \left (1-19 m-4 m^2+16 m^3\right )\right )+\frac{1}{8} a^3 (1+2 m) \left (B \left (13+12 m-16 m^2\right )+i A \left (37-52 m+16 m^2\right )\right ) \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right ) \tan ^{1+m}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(A-i B) \int \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}+\frac{\left ((1+2 m) \left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 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}{120 a^4}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right ) \tan ^{1+m}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(i A+B) \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 )}{8 a d}+\frac{\left ((1+2 m) \left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^m}{\sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{120 a^2 d}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right ) \tan ^{1+m}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left ((1+2 m) \left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 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 )}{120 a^2 d}+\frac{\left ((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 )}{8 a d \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{\left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right ) \tan ^{1+m}(c+d x)}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(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)}{8 a^2 d (1+m) \sqrt{a+i a \tan (c+d x)}}+\frac{(1+2 m) \left (B \left (13+12 m-16 m^2\right )+i A \left (37-52 m+16 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)}}{60 a^3 d}\\ \end{align*}

Mathematica [F]  time = 72.6787, size = 0, normalized size = 0. \[ \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.474, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2}{\left ({\left (A - i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (3 \, A - i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (3 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\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 (-5 i \, d x - 5 i \, c\right )}}{8 \, a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/8*sqrt(2)*((A - I*B)*e^(6*I*d*x + 6*I*c) + (3*A - I*B)*e^(4*I*d*x + 4*I*c) + (3*A + I*B)*e^(2*I*d*x
+ 2*I*c) + A + I*B)*((-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^(-5*I*d*x - 5*I*c)/a^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(tan(d*x+c)**m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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