3.11.23 \(\int \frac {1}{(a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}} \, dx\) [1023]

Optimal. Leaf size=186 \[ \frac {i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {8 \tan (e+f x)}{35 a^3 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

8/35*tan(f*x+e)/a^3/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1/7*I/f/(c-I*c*tan(f*x+e))^(1/2)/(a+I*
a*tan(f*x+e))^(7/2)+4/35*I/a/f/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2)+4/35*I/a^2/f/(c-I*c*tan(f*x+e
))^(1/2)/(a+I*a*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 47, 39} \begin {gather*} \frac {8 \tan (e+f x)}{35 a^3 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

(I/7)/(f*(a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)/35)/(a*f*(a + I*a*Tan[e + f*x])^(5/
2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)/35)/(a^2*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) +
(8*Tan[e + f*x])/(35*a^3*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

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 {1}{(a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{9/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(12 c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{35 a f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(8 c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{35 a^2 f}\\ &=\frac {i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {8 \tan (e+f x)}{35 a^3 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 2.23, size = 115, normalized size = 0.62 \begin {gather*} \frac {\sec ^2(e+f x) (-35 i-84 i \cos (2 (e+f x))+15 i \cos (4 (e+f x))+56 \sin (2 (e+f x))-20 \sin (4 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{280 a^3 c f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

(Sec[e + f*x]^2*(-35*I - (84*I)*Cos[2*(e + f*x)] + (15*I)*Cos[4*(e + f*x)] + 56*Sin[2*(e + f*x)] - 20*Sin[4*(e
+ f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(280*a^3*c*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.42, size = 130, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (24 i \left (\tan ^{5}\left (f x +e \right )\right )-8 \left (\tan ^{6}\left (f x +e \right )\right )+28 i \left (\tan ^{3}\left (f x +e \right )\right )+12 \left (\tan ^{4}\left (f x +e \right )\right )+4 i \tan \left (f x +e \right )+33 \left (\tan ^{2}\left (f x +e \right )\right )+13\right )}{35 f \,a^{4} c \left (-\tan \left (f x +e \right )+i\right )^{5} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(130\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (24 i \left (\tan ^{5}\left (f x +e \right )\right )-8 \left (\tan ^{6}\left (f x +e \right )\right )+28 i \left (\tan ^{3}\left (f x +e \right )\right )+12 \left (\tan ^{4}\left (f x +e \right )\right )+4 i \tan \left (f x +e \right )+33 \left (\tan ^{2}\left (f x +e \right )\right )+13\right )}{35 f \,a^{4} c \left (-\tan \left (f x +e \right )+i\right )^{5} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/35/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^4/c*(24*I*tan(f*x+e)^5-8*tan(f*x+e)^6+28*I*tan
(f*x+e)^3+12*tan(f*x+e)^4+4*I*tan(f*x+e)+33*tan(f*x+e)^2+13)/(-tan(f*x+e)+I)^5/(tan(f*x+e)+I)^2

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 0.79, size = 143, normalized size = 0.77 \begin {gather*} \frac {\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}} {\left (-35 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 208 i \, e^{\left (9 i \, f x + 9 i \, e\right )} + 105 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 208 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 210 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 98 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 33 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-7 i \, f x - 7 i \, e\right )}}{560 \, a^{4} c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/560*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-35*I*e^(10*I*f*x + 10*I*e) - 208*I
*e^(9*I*f*x + 9*I*e) + 105*I*e^(8*I*f*x + 8*I*e) - 208*I*e^(7*I*f*x + 7*I*e) + 210*I*e^(6*I*f*x + 6*I*e) + 98*
I*e^(4*I*f*x + 4*I*e) + 33*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-7*I*f*x - 7*I*e)/(a^4*c*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {7}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(7/2),x)

[Out]

Integral(1/((I*a*(tan(e + f*x) - I))**(7/2)*sqrt(-I*c*(tan(e + f*x) + I))), 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(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(7/2)*sqrt(-I*c*tan(f*x + e) + c)), x)

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Mupad [B]
time = 6.42, size = 183, normalized size = 0.98 \begin {gather*} \frac {\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}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,140{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,70{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,28{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,5{}\mathrm {i}+140\,\sin \left (2\,e+2\,f\,x\right )+70\,\sin \left (4\,e+4\,f\,x\right )+28\,\sin \left (6\,e+6\,f\,x\right )+5\,\sin \left (8\,e+8\,f\,x\right )-35{}\mathrm {i}\right )}{560\,a^4\,f\,\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a*(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)*140i + cos(
4*e + 4*f*x)*70i + cos(6*e + 6*f*x)*28i + cos(8*e + 8*f*x)*5i + 140*sin(2*e + 2*f*x) + 70*sin(4*e + 4*f*x) + 2
8*sin(6*e + 6*f*x) + 5*sin(8*e + 8*f*x) - 35i))/(560*a^4*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(
cos(2*e + 2*f*x) + 1))^(1/2))

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