3.809 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=227 \[ \frac {3 a^{5/2} (3 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {3 a^2 (3 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

3*a^(5/2)*(2*I*A+3*B)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f/c^(1/2)-3/2*
a^2*(2*I*A+3*B)*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c/f-1/2*a*(2*I*A+3*B)*(c-I*c*tan(f*x+e))^(1/
2)*(a+I*a*tan(f*x+e))^(3/2)/c/f-(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3588, 78, 50, 63, 217, 203} \[ \frac {3 a^{5/2} (3 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {3 a^2 (3 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

(3*a^(5/2)*((2*I)*A + 3*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/
(Sqrt[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (3*a^2*((2*I)*A + 3*B)
*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f) - (a*((2*I)*A + 3*B)*(a + I*a*Tan[e + f*x])^(3
/2)*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 203

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3588

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*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 11.25, size = 239, normalized size = 1.05 \[ \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \left (\frac {3 (3 B+2 i A) e^{-3 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}-\frac {(\tan (e+f x)+i) \sqrt {\sec (e+f x)} \sqrt {c-i c \tan (e+f x)} ((-5 B-2 i A) \sin (2 (e+f x))+(10 A-13 i B) \cos (2 (e+f x))+5 (2 A-3 i B))}{4 c}\right )}{f \sec ^{\frac {7}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x])*((3*((2*I)*A + 3*B)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e +
f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((3*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) - (Sqrt[Sec[e + f*x]
]*(5*(2*A - (3*I)*B) + (10*A - (13*I)*B)*Cos[2*(e + f*x)] + ((-2*I)*A - 5*B)*Sin[2*(e + f*x)])*(I + Tan[e + f*
x])*Sqrt[c - I*c*Tan[e + f*x]])/(4*c)))/(f*Sec[e + f*x]^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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fricas [B]  time = 0.82, size = 527, normalized size = 2.32 \[ -\frac {\sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 2 \, \sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (6 i \, A + 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A + 9 \, B\right )} a^{2}}\right ) - \sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 2 \, \sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (6 i \, A + 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A + 9 \, B\right )} a^{2}}\right ) - 2 \, {\left ({\left (-8 i \, A - 8 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-20 i \, A - 30 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-12 i \, A - 18 \, B\right )} a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt((36*A^2 - 108*I*A*B - 81*B^2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*log(2*(((24*I*A + 36*B)*
a^2*e^(3*I*f*x + 3*I*e) + (24*I*A + 36*B)*a^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*
I*f*x + 2*I*e) + 1)) + 2*sqrt((36*A^2 - 108*I*A*B - 81*B^2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((6*
I*A + 9*B)*a^2*e^(2*I*f*x + 2*I*e) + (6*I*A + 9*B)*a^2)) - sqrt((36*A^2 - 108*I*A*B - 81*B^2)*a^5/(c*f^2))*(c*
f*e^(2*I*f*x + 2*I*e) + c*f)*log(2*(((24*I*A + 36*B)*a^2*e^(3*I*f*x + 3*I*e) + (24*I*A + 36*B)*a^2*e^(I*f*x +
I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - 2*sqrt((36*A^2 - 108*I*A*B - 81*B^
2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((6*I*A + 9*B)*a^2*e^(2*I*f*x + 2*I*e) + (6*I*A + 9*B)*a^2))
- 2*((-8*I*A - 8*B)*a^2*e^(5*I*f*x + 5*I*e) + (-20*I*A - 30*B)*a^2*e^(3*I*f*x + 3*I*e) + (-12*I*A - 18*B)*a^2*
e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c*f*e^(2*I*f*x + 2*I*e)
+ c*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/sqrt(-I*c*tan(f*x + e) + c), x)

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maple [B]  time = 0.57, size = 565, normalized size = 2.49 \[ \frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (6 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +18 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +4 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )+9 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{3}\left (f x +e \right )\right )-6 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -12 i A \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-12 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -2 A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-14 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-9 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -19 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+10 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f c \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x)

[Out]

1/2*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c*(6*I*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*
x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c+18*I*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/
2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c+4*I*B*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2+9*B*ln
((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c-B*(c*a*(1+tan(f*x+e)^
2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^3-6*I*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1
/2))*a*c-12*I*A*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-12*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)
^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c-2*A*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-
14*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-9*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2
))/(c*a)^(1/2))*a*c-19*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+10*A*(c*a*(1+tan(f*x+e)^2))^(1/2)
*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^2/(c*a)^(1/2)

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maxima [B]  time = 1.10, size = 996, normalized size = 4.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-(16*(2*A - 7*I*B)*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-32*I*A - 112*B)*a^2*sin(3/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (24*(2*A - 3*I*B)*a^2*cos(4*f*x + 4*e) + 48*(2*A - 3*I*B)*a^2*cos
(2*f*x + 2*e) + (48*I*A + 72*B)*a^2*sin(4*f*x + 4*e) + (96*I*A + 144*B)*a^2*sin(2*f*x + 2*e) + 24*(2*A - 3*I*B
)*a^2)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + 1) - (24*(2*A - 3*I*B)*a^2*cos(4*f*x + 4*e) + 48*(2*A - 3*I*B)*a^2*cos(2*f*x + 2*e) + (48*I*A +
72*B)*a^2*sin(4*f*x + 4*e) + (96*I*A + 144*B)*a^2*sin(2*f*x + 2*e) + 24*(2*A - 3*I*B)*a^2)*arctan2(cos(1/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (64*(A
- I*B)*a^2*cos(4*f*x + 4*e) + 128*(A - I*B)*a^2*cos(2*f*x + 2*e) - (-64*I*A - 64*B)*a^2*sin(4*f*x + 4*e) - (-
128*I*A - 128*B)*a^2*sin(2*f*x + 2*e) + 48*(2*A - 3*I*B)*a^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) - ((24*I*A + 36*B)*a^2*cos(4*f*x + 4*e) + (48*I*A + 72*B)*a^2*cos(2*f*x + 2*e) - 12*(2*A - 3*I*B)*a^2*sin
(4*f*x + 4*e) - 24*(2*A - 3*I*B)*a^2*sin(2*f*x + 2*e) + (24*I*A + 36*B)*a^2)*log(cos(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) + 1) - ((-24*I*A - 36*B)*a^2*cos(4*f*x + 4*e) + (-48*I*A - 72*B)*a^2*cos(2*f*x +
2*e) + 12*(2*A - 3*I*B)*a^2*sin(4*f*x + 4*e) + 24*(2*A - 3*I*B)*a^2*sin(2*f*x + 2*e) + (-24*I*A - 36*B)*a^2)*
log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - ((-64*I*A - 64*B)*a^2*cos(4*f*x + 4*e) +
(-128*I*A - 128*B)*a^2*cos(2*f*x + 2*e) + 64*(A - I*B)*a^2*sin(4*f*x + 4*e) + 128*(A - I*B)*a^2*sin(2*f*x + 2
*e) + (-96*I*A - 144*B)*a^2)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-16*I*c*c
os(4*f*x + 4*e) - 32*I*c*cos(2*f*x + 2*e) + 16*c*sin(4*f*x + 4*e) + 32*c*sin(2*f*x + 2*e) - 16*I*c)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(5/2))/(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(5/2))/(c - c*tan(e + f*x)*1i)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

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

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