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

Optimal. Leaf size=246 \[ -\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 a^2 (46 B+45 i A) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{8 a^2 (46 B+45 i A) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{4 \sqrt{2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 a (59 B+60 i A) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

[Out]

(4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a^2*((45*I)*A + 46*
B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d) + (2*a^2*((45*I)*A + 46*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(
105*d) - (2*a^2*(3*A - (4*I)*B)*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(21*d) - (8*a*((60*I)*A + 59*B)*(a
+ I*a*Tan[c + d*x])^(3/2))/(315*d) + (((2*I)/9)*a*B*Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d

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Rubi [A]  time = 0.753477, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3594, 3597, 3592, 3527, 3480, 206} \[ -\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 a^2 (46 B+45 i A) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{8 a^2 (46 B+45 i A) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{4 \sqrt{2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 a (59 B+60 i A) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a^2*((45*I)*A + 46*
B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d) + (2*a^2*((45*I)*A + 46*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(
105*d) - (2*a^2*(3*A - (4*I)*B)*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(21*d) - (8*a*((60*I)*A + 59*B)*(a
+ I*a*Tan[c + d*x])^(3/2))/(315*d) + (((2*I)/9)*a*B*Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d

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 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 3592

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

Rule 3527

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

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \tan ^2(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 ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{2}{9} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac{3}{2} a (3 A-2 i B)+\frac{3}{2} a (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{4}{63} \int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{3}{4} a^2 (39 A-38 i B)+\frac{3}{4} a^2 (45 i A+46 B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{8 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{2} a^3 (45 i A+46 B)+\frac{3}{2} a^3 (60 A-59 i B) \tan (c+d x)\right ) \, dx}{315 a}\\ &=\frac{2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}-\frac{8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{8 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{2} a^3 (60 A-59 i B)-\frac{3}{2} a^3 (45 i A+46 B) \tan (c+d x)\right ) \, dx}{315 a}\\ &=-\frac{8 a^2 (45 i A+46 B) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}-\frac{8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}-\left (4 a^2 (A-i B)\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{8 a^2 (45 i A+46 B) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}-\frac{8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac{\left (8 a^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt{2} a^{5/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 a^2 (45 i A+46 B) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{2 a^2 (45 i A+46 B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{2 a^2 (3 A-4 i B) \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{21 d}-\frac{8 a (60 i A+59 B) (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{2 i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}\\ \end{align*}

Mathematica [A]  time = 5.6426, size = 284, normalized size = 1.15 \[ \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (4 \sqrt{2} (B+i A) e^{-3 i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac{i (\cos (2 c)-i \sin (2 c)) \sec ^{\frac{9}{2}}(c+d x) (12 (260 A-251 i B) \cos (2 (c+d x))+(915 A-961 i B) \cos (4 (c+d x))+390 i A \sin (2 (c+d x))+285 i A \sin (4 (c+d x))+2205 A+282 B \sin (2 (c+d x))+331 B \sin (4 (c+d x))-2331 i B)}{1260 (\cos (d x)+i \sin (d x))^2}\right )}{d \sec ^{\frac{7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*Sqrt[2]*(I*A + B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E
^(I*(c + d*x))])/E^((3*I)*(c + d*x)) - ((I/1260)*Sec[c + d*x]^(9/2)*(Cos[2*c] - I*Sin[2*c])*(2205*A - (2331*I)
*B + 12*(260*A - (251*I)*B)*Cos[2*(c + d*x)] + (915*A - (961*I)*B)*Cos[4*(c + d*x)] + (390*I)*A*Sin[2*(c + d*x
)] + 282*B*Sin[2*(c + d*x)] + (285*I)*A*Sin[4*(c + d*x)] + 331*B*Sin[4*(c + d*x)]))/(Cos[d*x] + I*Sin[d*x])^2)
*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(d*Sec[c + d*x]^(7/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [A]  time = 0.027, size = 206, normalized size = 0.8 \begin{align*}{\frac{-2\,i}{{a}^{2}d} \left ( -{\frac{i}{9}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{9}{2}}}+{\frac{i}{7}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}a+{\frac{Aa}{7} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{i}{5}}B{a}^{2} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}-{\frac{i}{3}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}{a}^{3}+{\frac{A{a}^{3}}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-2\,iB{a}^{4}\sqrt{a+ia\tan \left ( dx+c \right ) }+2\,A{a}^{4}\sqrt{a+ia\tan \left ( dx+c \right ) }-2\,{a}^{9/2} \left ( A-iB \right ) \sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/d/a^2*(-1/9*I*B*(a+I*a*tan(d*x+c))^(9/2)+1/7*I*B*(a+I*a*tan(d*x+c))^(7/2)*a+1/7*A*(a+I*a*tan(d*x+c))^(7/2
)*a-1/5*I*B*a^2*(a+I*a*tan(d*x+c))^(5/2)-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)*a^3+1/3*A*(a+I*a*tan(d*x+c))^(3/2)*a
^3-2*I*B*a^4*(a+I*a*tan(d*x+c))^(1/2)+2*A*a^4*(a+I*a*tan(d*x+c))^(1/2)-2*a^(9/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(
a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.79612, size = 1585, normalized size = 6.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/630*(sqrt(2)*((-4800*I*A - 5168*B)*a^2*e^(8*I*d*x + 8*I*c) + (-14040*I*A - 13176*B)*a^2*e^(6*I*d*x + 6*I*c)
+ (-17640*I*A - 18648*B)*a^2*e^(4*I*d*x + 4*I*c) + (-10920*I*A - 10920*B)*a^2*e^(2*I*d*x + 2*I*c) + (-2520*I*A
- 2520*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 315*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^5
/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d
)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e
^(I*d*x + I*c) + sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I
*A + 4*B)*a^2)) - 315*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*
I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I
*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - sqrt(-(32*A^2 - 64*I*A*B - 32*B^2
)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6
*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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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)**2*(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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(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)^2, x)