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

Optimal. Leaf size=141 \[ -\frac {4 \sqrt {2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (B+i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (B+i A) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3527, 3478, 3480, 206} \[ \frac {4 a^2 (B+i A) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 \sqrt {2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (B+i A) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(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 + (4*a^2*(I*A + B)*Sqrt
[a + I*a*Tan[c + d*x]])/d + (2*a*(I*A + B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d) + (2*B*(a + I*a*Tan[c + d*x])^(
5/2))/(5*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))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}-(-A+i B) \int (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}+(2 a (A-i B)) \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (4 a^2 (A-i B)\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 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 {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end {align*}

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Mathematica [A]  time = 3.46, size = 236, normalized size = 1.67 \[ \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (\frac {(\cos (2 c)-i \sin (2 c)) \sec ^{\frac {5}{2}}(c+d x) ((-5 A+11 i B) \sin (2 (c+d x))+(41 B+35 i A) \cos (2 (c+d x))+35 (B+i A))}{15 (\cos (d x)+i \sin (d x))^2}-4 i \sqrt {2} (A-i B) e^{-3 i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((((-4*I)*Sqrt[2]*(A - I*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcS
inh[E^(I*(c + d*x))])/E^((3*I)*(c + d*x)) + (Sec[c + d*x]^(5/2)*(Cos[2*c] - I*Sin[2*c])*(35*(I*A + B) + ((35*I
)*A + 41*B)*Cos[2*(c + d*x)] + (-5*A + (11*I)*B)*Sin[2*(c + d*x)]))/(15*(Cos[d*x] + I*Sin[d*x])^2))*(a + I*a*T
an[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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fricas [B]  time = 0.86, size = 424, normalized size = 3.01 \[ -\frac {15 \, \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (16 i \, A + 16 \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) - 15 \, \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left ({\left (16 i \, A + 16 \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {-\frac {{\left (128 \, A^{2} - 256 i \, A B - 128 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) - \sqrt {2} {\left ({\left (320 i \, A + 416 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (560 i \, A + 560 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (240 i \, A + 240 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
*log(((16*I*A + 16*B)*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/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)/((4*I*A + 4*B)*a^2)) - 15*sqrt(-(128*A^2
- 256*I*A*B - 128*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(((16*I*A + 16*B)*a^3
*e^(I*d*x + I*c) - sqrt(2)*sqrt(-(128*A^2 - 256*I*A*B - 128*B^2)*a^5/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)/((4*I*A + 4*B)*a^2)) - sqrt(2)*((320*I*A + 416*B)*a^2*e^(5*I*d*x +
5*I*c) + (560*I*A + 560*B)*a^2*e^(3*I*d*x + 3*I*c) + (240*I*A + 240*B)*a^2*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))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.19, size = 141, normalized size = 1.00 \[ \frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}+\frac {A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 a^{2} A \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/d*(-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-2
*I*B*a^2*(a+I*a*tan(d*x+c))^(1/2)+2*a^2*A*(a+I*a*tan(d*x+c))^(1/2)-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 = 0.95, size = 134, normalized size = 0.95 \[ \frac {2 i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 3 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{2} + 30 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{3}\right )}}{15 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*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) + sq
rt(I*a*tan(d*x + c) + a))) - 3*I*(I*a*tan(d*x + c) + a)^(5/2)*B*a + 5*(I*a*tan(d*x + c) + a)^(3/2)*(A - I*B)*a
^2 + 30*sqrt(I*a*tan(d*x + c) + a)*(A - I*B)*a^3)/(a*d)

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mupad [B]  time = 0.96, size = 188, normalized size = 1.33 \[ \frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}+\frac {A\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {2\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {A\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}+\frac {4\,B\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,A\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d}+\frac {\sqrt {2}\,B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \]

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)^(5/2),x)

[Out]

(2*B*(a + a*tan(c + d*x)*1i)^(5/2))/(5*d) + (A*a*(a + a*tan(c + d*x)*1i)^(3/2)*2i)/(3*d) + (2*B*a*(a + a*tan(c
+ d*x)*1i)^(3/2))/(3*d) + (A*a^2*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/d + (4*B*a^2*(a + a*tan(c + d*x)*1i)^(1/2)
)/d - (2^(1/2)*A*(-a)^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*4i)/d + (2^(1/2)*B*a^
(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*4i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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