3.74 \(\int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=146 \[ -\frac {a^2}{32 d (a \sin (c+d x)+a)^4}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac {a}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {3}{32 d (a \sin (c+d x)+a)^2} \]

[Out]

15/64*arctanh(sin(d*x+c))/a^2/d+1/64/d/(a-a*sin(d*x+c))^2-1/32*a^2/d/(a+a*sin(d*x+c))^4-1/16*a/d/(a+a*sin(d*x+
c))^3-3/32/d/(a+a*sin(d*x+c))^2+5/64/d/(a^2-a^2*sin(d*x+c))-5/32/d/(a^2+a^2*sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2667, 44, 206} \[ -\frac {a^2}{32 d (a \sin (c+d x)+a)^4}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac {a}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {3}{32 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sin[c + d*x])^2,x]

[Out]

(15*ArcTanh[Sin[c + d*x]])/(64*a^2*d) + 1/(64*d*(a - a*Sin[c + d*x])^2) - a^2/(32*d*(a + a*Sin[c + d*x])^4) -
a/(16*d*(a + a*Sin[c + d*x])^3) - 3/(32*d*(a + a*Sin[c + d*x])^2) + 5/(64*d*(a^2 - a^2*Sin[c + d*x])) - 5/(32*
d*(a^2 + a^2*Sin[c + d*x]))

Rule 44

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

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 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^3}+\frac {5}{64 a^6 (a-x)^2}+\frac {1}{8 a^3 (a+x)^5}+\frac {3}{16 a^4 (a+x)^4}+\frac {3}{16 a^5 (a+x)^3}+\frac {5}{32 a^6 (a+x)^2}+\frac {15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d}\\ &=\frac {15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 137, normalized size = 0.94 \[ \frac {(1-\sin (c+d x))^2 (\sin (c+d x)+1)^2 \sec ^4(c+d x) \left (\frac {5}{64 (1-\sin (c+d x))}-\frac {5}{32 (\sin (c+d x)+1)}+\frac {1}{64 (1-\sin (c+d x))^2}-\frac {3}{32 (\sin (c+d x)+1)^2}-\frac {1}{16 (\sin (c+d x)+1)^3}-\frac {1}{32 (\sin (c+d x)+1)^4}+\frac {15}{64} \tanh ^{-1}(\sin (c+d x))\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^4*(1 - Sin[c + d*x])^2*(1 + Sin[c + d*x])^2*((15*ArcTanh[Sin[c + d*x]])/64 + 1/(64*(1 - Sin[c +
d*x])^2) + 5/(64*(1 - Sin[c + d*x])) - 1/(32*(1 + Sin[c + d*x])^4) - 1/(16*(1 + Sin[c + d*x])^3) - 3/(32*(1 +
Sin[c + d*x])^2) - 5/(32*(1 + Sin[c + d*x]))))/(a^2*d)

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fricas [A]  time = 0.85, size = 198, normalized size = 1.36 \[ \frac {60 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) - 8}{128 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/128*(60*cos(d*x + c)^4 - 20*cos(d*x + c)^2 + 15*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x
+ c)^4)*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(-si
n(d*x + c) + 1) + 2*(15*cos(d*x + c)^4 - 20*cos(d*x + c)^2 - 12)*sin(d*x + c) - 8)/(a^2*d*cos(d*x + c)^6 - 2*a
^2*d*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^4)

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giac [A]  time = 0.54, size = 126, normalized size = 0.86 \[ \frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (45 \, \sin \left (d x + c\right )^{2} - 110 \, \sin \left (d x + c\right ) + 69\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 580 \, \sin \left (d x + c\right )^{3} + 1038 \, \sin \left (d x + c\right )^{2} + 868 \, \sin \left (d x + c\right ) + 301}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/512*(60*log(abs(sin(d*x + c) + 1))/a^2 - 60*log(abs(sin(d*x + c) - 1))/a^2 + 2*(45*sin(d*x + c)^2 - 110*sin(
d*x + c) + 69)/(a^2*(sin(d*x + c) - 1)^2) - (125*sin(d*x + c)^4 + 580*sin(d*x + c)^3 + 1038*sin(d*x + c)^2 + 8
68*sin(d*x + c) + 301)/(a^2*(sin(d*x + c) + 1)^4))/d

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maple [A]  time = 0.23, size = 144, normalized size = 0.99 \[ \frac {1}{64 a^{2} d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{64 a^{2} d \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right )}{128 a^{2} d}-\frac {1}{32 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 a^{2} d \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right )}{128 a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x)

[Out]

1/64/a^2/d/(sin(d*x+c)-1)^2-5/64/a^2/d/(sin(d*x+c)-1)-15/128/a^2/d*ln(sin(d*x+c)-1)-1/32/a^2/d/(1+sin(d*x+c))^
4-1/16/a^2/d/(1+sin(d*x+c))^3-3/32/a^2/d/(1+sin(d*x+c))^2-5/32/a^2/d/(1+sin(d*x+c))+15/128*ln(1+sin(d*x+c))/a^
2/d

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maxima [A]  time = 0.37, size = 167, normalized size = 1.14 \[ -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} + 30 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{3} - 50 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/128*(2*(15*sin(d*x + c)^5 + 30*sin(d*x + c)^4 - 10*sin(d*x + c)^3 - 50*sin(d*x + c)^2 - 17*sin(d*x + c) + 1
6)/(a^2*sin(d*x + c)^6 + 2*a^2*sin(d*x + c)^5 - a^2*sin(d*x + c)^4 - 4*a^2*sin(d*x + c)^3 - a^2*sin(d*x + c)^2
+ 2*a^2*sin(d*x + c) + a^2) - 15*log(sin(d*x + c) + 1)/a^2 + 15*log(sin(d*x + c) - 1)/a^2)/d

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mupad [B]  time = 0.19, size = 151, normalized size = 1.03 \[ \frac {15\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{64\,a^2\,d}+\frac {-\frac {15\,{\sin \left (c+d\,x\right )}^5}{64}-\frac {15\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {25\,{\sin \left (c+d\,x\right )}^2}{32}+\frac {17\,\sin \left (c+d\,x\right )}{64}-\frac {1}{4}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^6+2\,a^2\,{\sin \left (c+d\,x\right )}^5-a^2\,{\sin \left (c+d\,x\right )}^4-4\,a^2\,{\sin \left (c+d\,x\right )}^3-a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a*sin(c + d*x))^2),x)

[Out]

(15*atanh(sin(c + d*x)))/(64*a^2*d) + ((17*sin(c + d*x))/64 + (25*sin(c + d*x)^2)/32 + (5*sin(c + d*x)^3)/32 -
(15*sin(c + d*x)^4)/32 - (15*sin(c + d*x)^5)/64 - 1/4)/(d*(2*a^2*sin(c + d*x) + a^2 - a^2*sin(c + d*x)^2 - 4*
a^2*sin(c + d*x)^3 - a^2*sin(c + d*x)^4 + 2*a^2*sin(c + d*x)^5 + a^2*sin(c + d*x)^6))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**5/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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