3.68 \(\int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=143 \[ -\frac {2 (B+5 i A) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {8 B \sqrt {a+i a \tan (c+d x)}}{5 d} \]

[Out]

(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d-8/5*B*(a+I*a*tan(d*x+c))^(1/2)
/d+2/5*B*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d-2/15*(5*I*A+B)*(a+I*a*tan(d*x+c))^(3/2)/a/d

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Rubi [A]  time = 0.30, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3597, 3592, 3527, 3480, 206} \[ -\frac {2 (B+5 i A) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {8 B \sqrt {a+i a \tan (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*Sqrt[a]*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*B*Sqrt[a + I*a*Tan[c
+ d*x]])/(5*d) + (2*B*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*((5*I)*A + B)*(a + I*a*Tan[c + d*x
])^(3/2))/(15*a*d)

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])

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 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 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 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]

Rubi steps

\begin {align*} \int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-2 a B+\frac {1}{2} a (5 A-i B) \tan (c+d x)\right ) \, dx}{5 a}\\ &=\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac {2 \int \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{2} a (5 A-i B)-2 a B \tan (c+d x)\right ) \, dx}{5 a}\\ &=-\frac {8 B \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}+(-A+i B) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {8 B \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac {(2 a (i A+B)) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {\sqrt {2} \sqrt {a} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 B \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (5 i A+B) (a+i a \tan (c+d x))^{3/2}}{15 a d}\\ \end {align*}

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Mathematica [A]  time = 2.90, size = 184, normalized size = 1.29 \[ \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \left (\frac {\sqrt {2} (B+i A) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}}}+\frac {2}{15} \sqrt {\sec (c+d x)} \left ((5 A-i B) \tan (c+d x)-5 i A+3 B \sec ^2(c+d x)-16 B\right )\right )}{d \sec ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x])*((Sqrt[2]*(I*A + B)*ArcSinh[E^(I*(c + d*x))])/(Sqrt[E^(I*(c +
d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) + (2*Sqrt[Sec[c + d*x]]*((-5*I)*A - 16*B + 3*
B*Sec[c + d*x]^2 + (5*A - I*B)*Tan[c + d*x]))/15))/(d*Sec[c + d*x]^(3/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.50, size = 387, normalized size = 2.71 \[ \frac {15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (-80 i \, A - 136 \, B\right )} e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (-80 i \, A - 160 \, B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} - 120 \, B e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*log(((4
*I*A + 4*B)*a*e^(I*d*x + I*c) + sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*
I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*log(((4*I*A + 4*B)*a*e^(I*d*x + I*c) - sqrt(2)*(d*e^(2*I*d*x
+ 2*I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*
A + B)) + sqrt(2)*((-80*I*A - 136*B)*e^(5*I*d*x + 5*I*c) + (-80*I*A - 160*B)*e^(3*I*d*x + 3*I*c) - 120*B*e^(I*
d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.35, size = 124, normalized size = 0.87 \[ -\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}+\frac {A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}-i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/d/a^2*(-1/5*I*B*(a+I*a*tan(d*x+c))^(5/2)+1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)*a+1/3*A*(a+I*a*tan(d*x+c))^(3/2
)*a-I*a^2*B*(a+I*a*tan(d*x+c))^(1/2)-1/2*a^(5/2)*(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 [A]  time = 0.79, size = 130, normalized size = 0.91 \[ -\frac {i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 12 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A + i \, B\right )} a^{2} - 60 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a^{3}\right )}}{30 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*I*(15*sqrt(2)*(A - I*B)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + s
qrt(I*a*tan(d*x + c) + a))) - 12*I*(I*a*tan(d*x + c) + a)^(5/2)*B*a + 20*(I*a*tan(d*x + c) + a)^(3/2)*(A + I*B
)*a^2 - 60*I*sqrt(I*a*tan(d*x + c) + a)*B*a^3)/(a^3*d)

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mupad [B]  time = 7.12, size = 168, normalized size = 1.17 \[ -\frac {2\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a\,d}+\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a\,d}-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^2\,d}+\frac {\sqrt {2}\,A\,\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{d}-\frac {\sqrt {2}\,B\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a*d) - (A*(a + a*tan(c + d*x)*1i)^(3/2)*2i)/(3*a*d) - (2*B*(a + a*tan(c
+ d*x)*1i)^(1/2))/d - (2*B*(a + a*tan(c + d*x)*1i)^(5/2))/(5*a^2*d) + (2^(1/2)*A*(-a)^(1/2)*atan((2^(1/2)*(a
+ a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/d - (2^(1/2)*B*a^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(
1/2)*1i)/(2*a^(1/2)))*1i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*(A + B*tan(c + d*x))*tan(c + d*x)**2, x)

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