3.77 \(\int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 \sqrt {2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (B+i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3527, 3478, 3480, 206} \[ -\frac {2 \sqrt {2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (B+i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

(-2*Sqrt[2]*a^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (2*a*(I*A + B)*Sqrt[a
+ I*a*Tan[c + d*x]])/d + (2*B*(a + I*a*Tan[c + d*x])^(3/2))/(3*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 3478

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

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]

Rubi steps

\begin {align*} \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d}-(-A+i B) \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {2 a (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d}+(2 a (A-i B)) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {2 a (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {\left (4 a^2 (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 {2 \sqrt {2} a^{3/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 2.85, size = 190, normalized size = 1.78 \[ \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac {2}{3} (\cos (c)-i \sin (c)) \sqrt {\sec (c+d x)} (\sin (d x)+i \cos (d x)) (3 A+B \tan (c+d x)-4 i B)-\frac {2 i \sqrt {2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}\right )}{d \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*(((-2*I)*Sqrt[2]*(A - I*B)*ArcSinh[E^(I*(c + d*x))])/((E^(I
*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) + (2*Sqrt[Sec[c + d*x]]*(Cos[c]
- I*Sin[c])*(I*Cos[d*x] + Sin[d*x])*(3*A - (4*I)*B + B*Tan[c + d*x]))/3))/(d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x
] + B*Sin[c + d*x]))

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fricas [B]  time = 0.64, size = 364, normalized size = 3.40 \[ -\frac {3 \, \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (8 i \, A + 8 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) - 3 \, \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (8 i \, A + 8 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) - \sqrt {2} {\left ({\left (24 i \, A + 40 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (24 i \, A + 24 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(((8*I*A + 8*B)*a^2*e^(I*d
*x + I*c) + sqrt(2)*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) - 3*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(d*e^(2*I
*d*x + 2*I*c) + d)*log(((8*I*A + 8*B)*a^2*e^(I*d*x + I*c) - sqrt(2)*sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2
)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) - sqrt(2)
*((24*I*A + 40*B)*a*e^(3*I*d*x + 3*I*c) + (24*I*A + 24*B)*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))
)/(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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.19, size = 99, normalized size = 0.93 \[ \frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-i B a \sqrt {a +i a \tan \left (d x +c \right )}+A \sqrt {a +i a \tan \left (d x +c \right )}\, a -a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

2*I/d*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)-I*B*a*(a+I*a*tan(d*x+c))^(1/2)+A*(a+I*a*tan(d*x+c))^(1/2)*a-a^(3/2)*(
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 = 0.69, size = 111, normalized size = 1.04 \[ \frac {i \, {\left (3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 2 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 6 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{2}\right )}}{3 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/3*I*(3*sqrt(2)*(A - I*B)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt
(I*a*tan(d*x + c) + a))) - 2*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a + 6*sqrt(I*a*tan(d*x + c) + a)*(A - I*B)*a^2)/
(a*d)

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mupad [B]  time = 6.64, size = 139, normalized size = 1.30 \[ \frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {A\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {\sqrt {2}\,A\,{\left (-a\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,2{}\mathrm {i}}{d}-\frac {2\,\sqrt {2}\,B\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x)), x)

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