3.11.5 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx\) [1005]

Optimal. Leaf size=136 \[ \frac {i (c-i c \tan (e+f x))^{3/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

1/7*I*(c-I*c*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(7/2)+2/35*I*(c-I*c*tan(f*x+e))^(3/2)/a/f/(a+I*a*tan(f*x+e
))^(5/2)+2/105*I*(c-I*c*tan(f*x+e))^(3/2)/a^2/f/(a+I*a*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 47, 37} \begin {gather*} \frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{7 f (a+i a \tan (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(7/2),x]

[Out]

((I/7)*(c - I*c*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^(7/2)) + (((2*I)/35)*(c - I*c*Tan[e + f*x])^(3/
2))/(a*f*(a + I*a*Tan[e + f*x])^(5/2)) + (((2*I)/105)*(c - I*c*Tan[e + f*x])^(3/2))/(a^2*f*(a + I*a*Tan[e + f*
x])^(3/2))

Rule 37

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

Rule 47

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

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((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), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {(2 c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{35 a f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 2.17, size = 101, normalized size = 0.74 \begin {gather*} \frac {i c \sec ^2(e+f x) (21+25 \cos (2 (e+f x))+10 i \sin (2 (e+f x))) (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{210 a^3 f (-i+\tan (e+f x))^3 \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(7/2),x]

[Out]

((I/210)*c*Sec[e + f*x]^2*(21 + 25*Cos[2*(e + f*x)] + (10*I)*Sin[2*(e + f*x)])*(I + Tan[e + f*x])*Sqrt[c - I*c
*Tan[e + f*x]])/(a^3*f*(-I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.37, size = 85, normalized size = 0.62

method	result	size
derivativedivides	\(-\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (10 i \tan \left (f x +e \right )-2 \left (\tan ^{2}\left (f x +e \right )\right )+23\right )}{105 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}}\)	\(85\)
default	\(-\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (10 i \tan \left (f x +e \right )-2 \left (\tan ^{2}\left (f x +e \right )\right )+23\right )}{105 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/105/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^4*c*(1+tan(f*x+e)^2)*(10*I*tan(f*x+e)-2*tan(
f*x+e)^2+23)/(-tan(f*x+e)+I)^5

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Maxima [A]
time = 0.63, size = 146, normalized size = 1.07 \begin {gather*} \frac {{\left (15 i \, c \cos \left (7 \, f x + 7 \, e\right ) + 42 i \, c \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right ) + 35 i \, c \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right ) + 15 \, c \sin \left (7 \, f x + 7 \, e\right ) + 42 \, c \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right ) + 35 \, c \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right )\right )} \sqrt {c}}{420 \, a^{\frac {7}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

1/420*(15*I*c*cos(7*f*x + 7*e) + 42*I*c*cos(5/7*arctan2(sin(7*f*x + 7*e), cos(7*f*x + 7*e))) + 35*I*c*cos(3/7*
arctan2(sin(7*f*x + 7*e), cos(7*f*x + 7*e))) + 15*c*sin(7*f*x + 7*e) + 42*c*sin(5/7*arctan2(sin(7*f*x + 7*e),
cos(7*f*x + 7*e))) + 35*c*sin(3/7*arctan2(sin(7*f*x + 7*e), cos(7*f*x + 7*e))))*sqrt(c)/(a^(7/2)*f)

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Fricas [A]
time = 1.43, size = 97, normalized size = 0.71 \begin {gather*} \frac {{\left (35 i \, c e^{\left (6 i \, f x + 6 i \, e\right )} + 77 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 57 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i \, c\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}} e^{\left (-7 i \, f x - 7 i \, e\right )}}{420 \, a^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/420*(35*I*c*e^(6*I*f*x + 6*I*e) + 77*I*c*e^(4*I*f*x + 4*I*e) + 57*I*c*e^(2*I*f*x + 2*I*e) + 15*I*c)*sqrt(a/(
e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-7*I*f*x - 7*I*e)/(a^4*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**(7/2),x)

[Out]

Integral((-I*c*(tan(e + f*x) + I))**(3/2)/(I*a*(tan(e + f*x) - I))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a)^(7/2), x)

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Mupad [B]
time = 6.51, size = 182, normalized size = 1.34 \begin {gather*} \frac {c\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (35\,\sin \left (2\,e+2\,f\,x\right )+77\,\sin \left (4\,e+4\,f\,x\right )+57\,\sin \left (6\,e+6\,f\,x\right )+15\,\sin \left (8\,e+8\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,35{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,77{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,57{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,15{}\mathrm {i}\right )}{840\,a^4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*tan(e + f*x)*1i)^(3/2)/(a + a*tan(e + f*x)*1i)^(7/2),x)

[Out]

(c*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*((c*(cos(2*e + 2*f*x) - sin
(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*35i + cos(4*e + 4*f*x)*77i + cos(6*e +
6*f*x)*57i + cos(8*e + 8*f*x)*15i + 35*sin(2*e + 2*f*x) + 77*sin(4*e + 4*f*x) + 57*sin(6*e + 6*f*x) + 15*sin(8
*e + 8*f*x)))/(840*a^4*f)

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