3.296 \(\int \frac{1}{(a+b \sec ^2(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 d (a+b)^3 \sqrt{a+b \tan ^2(c+d x)+b}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)+b}}\right )}{a^{7/2} d}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 d (a+b)^2 \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}-\frac{b \tan (c+d x)}{5 a d (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}} \]

[Out]

ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + b + b*Tan[c + d*x]^2]]/(a^(7/2)*d) - (b*Tan[c + d*x])/(5*a*(a + b)*d*(a
+ b + b*Tan[c + d*x]^2)^(5/2)) - (b*(9*a + 5*b)*Tan[c + d*x])/(15*a^2*(a + b)^2*d*(a + b + b*Tan[c + d*x]^2)^
(3/2)) - (b*(33*a^2 + 40*a*b + 15*b^2)*Tan[c + d*x])/(15*a^3*(a + b)^3*d*Sqrt[a + b + b*Tan[c + d*x]^2])

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Rubi [A]  time = 0.193807, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4128, 414, 527, 12, 377, 203} \[ -\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 d (a+b)^3 \sqrt{a+b \tan ^2(c+d x)+b}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)+b}}\right )}{a^{7/2} d}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 d (a+b)^2 \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}-\frac{b \tan (c+d x)}{5 a d (a+b) \left (a+b \tan ^2(c+d x)+b\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x]^2)^(-7/2),x]

[Out]

ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + b + b*Tan[c + d*x]^2]]/(a^(7/2)*d) - (b*Tan[c + d*x])/(5*a*(a + b)*d*(a
+ b + b*Tan[c + d*x]^2)^(5/2)) - (b*(9*a + 5*b)*Tan[c + d*x])/(15*a^2*(a + b)^2*d*(a + b + b*Tan[c + d*x]^2)^
(3/2)) - (b*(33*a^2 + 40*a*b + 15*b^2)*Tan[c + d*x])/(15*a^3*(a + b)^3*d*Sqrt[a + b + b*Tan[c + d*x]^2])

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sec ^2(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{7/2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{5 a+b-4 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{5 a (a+b) d}\\ &=-\frac{b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{15 a^2+12 a b+5 b^2-2 b (9 a+5 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{15 a^2 (a+b)^2 d}\\ &=-\frac{b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b+b \tan ^2(c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{15 (a+b)^3}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (c+d x)\right )}{15 a^3 (a+b)^3 d}\\ &=-\frac{b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b+b \tan ^2(c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=-\frac{b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b+b \tan ^2(c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (c+d x)}{\sqrt{a+b+b \tan ^2(c+d x)}}\right )}{a^3 d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+b+b \tan ^2(c+d x)}}\right )}{a^{7/2} d}-\frac{b \tan (c+d x)}{5 a (a+b) d \left (a+b+b \tan ^2(c+d x)\right )^{5/2}}-\frac{b (9 a+5 b) \tan (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}-\frac{b \left (33 a^2+40 a b+15 b^2\right ) \tan (c+d x)}{15 a^3 (a+b)^3 d \sqrt{a+b+b \tan ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 18.7578, size = 1777, normalized size = 9.93 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sec[c + d*x]^2)^(-7/2),x]

[Out]

(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^6*Sin[c + d*x]
)/(8*Sqrt[2]*d*(a + b*Sec[c + d*x]^2)^(7/2)*(a + b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2,
3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a*AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c
+ d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c
+ d*x]^2)*((21*a*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^
7*Sin[c + d*x]^2)/(8*Sqrt[2]*(a + b - a*Sin[c + d*x]^2)^(9/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d
*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a*AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a +
b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d*x]^2)) + (3
*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^7)/(8*Sqrt[2]*(a
+ b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a +
b)] + (7*a*AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -
2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d*x]^2)) - (9*(a + b)*AppellF1[1/2, -3, 7/2,
3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^5*Sin[c + d*x]^2)/(4*Sqrt[2]*(a + b - a*Sin[c +
d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a*Appel
lF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin
[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d*x]^2)) + (3*(a + b)*Cos[c + d*x]^6*Sin[c + d*x]*((7*a*d*Ap
pellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[c + d*x])/(3*(a + b)) -
2*d*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[c + d*x]))/(8*Sq
rt[2]*d*(a + b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x
]^2)/(a + b)] + (7*a*AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*Appel
lF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d*x]^2)) - (3*(a + b)*AppellF1[1/2
, -3, 7/2, 3/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]^6*Sin[c + d*x]*(2*d*(7*a*AppellF1[3/2
, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*
x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Cos[c + d*x]*Sin[c + d*x] + 3*(a + b)*((7*a*d*AppellF1[3/2, -3, 9/2, 5/2, S
in[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[c + d*x])/(3*(a + b)) - 2*d*AppellF1[3/2, -2, 7/2,
5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[c + d*x]) + Sin[c + d*x]^2*(7*a*((27*a*d*Ap
pellF1[5/2, -3, 11/2, 7/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[c + d*x])/(5*(a + b))
- (18*d*AppellF1[5/2, -2, 9/2, 7/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[c + d*x])/5)
- 6*(a + b)*((21*a*d*AppellF1[5/2, -2, 9/2, 7/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c + d*x]*Sin[
c + d*x])/(5*(a + b)) - (12*d*AppellF1[5/2, -1, 7/2, 7/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)]*Cos[c +
d*x]*Sin[c + d*x])/5))))/(8*Sqrt[2]*d*(a + b - a*Sin[c + d*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -3, 7/2, 3/2,
Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)] + (7*a*AppellF1[3/2, -3, 9/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]
^2)/(a + b)] - 6*(a + b)*AppellF1[3/2, -2, 7/2, 5/2, Sin[c + d*x]^2, (a*Sin[c + d*x]^2)/(a + b)])*Sin[c + d*x]
^2)^2)))

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Maple [C]  time = 0.862, size = 6116, normalized size = 34.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sec(d*x+c)^2)^(7/2),x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c)^2)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.40266, size = 2795, normalized size = 15.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c)^2)^(7/2),x, algorithm="fricas")

[Out]

[-1/120*(15*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(d*x + c)^6 + a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6 + 3*(a
^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*cos(d*x + c)^4 + 3*(a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*cos(d*x +
c)^2)*sqrt(-a)*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 - a^3*b)*cos(d*x + c)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b
^2)*cos(d*x + c)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos
(d*x + c)^2 + 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 - a^2*b)*cos(d*x + c)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(
d*x + c)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2
)*sin(d*x + c)) + 8*((45*a^5*b + 60*a^4*b^2 + 23*a^3*b^3)*cos(d*x + c)^5 + (75*a^4*b^2 + 94*a^3*b^3 + 35*a^2*b
^4)*cos(d*x + c)^3 + (33*a^3*b^3 + 40*a^2*b^4 + 15*a*b^5)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x +
c)^2)*sin(d*x + c))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7*b^3)*d*cos(d*x + c)^6 + 3*(a^9*b + 3*a^8*b^2 + 3*a^7*b^
3 + a^6*b^4)*d*cos(d*x + c)^4 + 3*(a^8*b^2 + 3*a^7*b^3 + 3*a^6*b^4 + a^5*b^5)*d*cos(d*x + c)^2 + (a^7*b^3 + 3*
a^6*b^4 + 3*a^5*b^5 + a^4*b^6)*d), -1/60*(15*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(d*x + c)^6 + a^3*b^3 +
3*a^2*b^4 + 3*a*b^5 + b^6 + 3*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*cos(d*x + c)^4 + 3*(a^4*b^2 + 3*a^3*b
^3 + 3*a^2*b^4 + a*b^5)*cos(d*x + c)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(d*x + c)^5 - 8*(a^2 - a*b)*cos(d*x + c)^
3 + (a^2 - 6*a*b + b^2)*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)/((2*a^3*cos(d*x + c)
^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(d*x + c)^2)*sin(d*x + c))) + 4*((45*a^5*b + 60*a^4*b^2 + 23*a^3*b^3)*
cos(d*x + c)^5 + (75*a^4*b^2 + 94*a^3*b^3 + 35*a^2*b^4)*cos(d*x + c)^3 + (33*a^3*b^3 + 40*a^2*b^4 + 15*a*b^5)*
cos(d*x + c))*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)*sin(d*x + c))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7*b^3
)*d*cos(d*x + c)^6 + 3*(a^9*b + 3*a^8*b^2 + 3*a^7*b^3 + a^6*b^4)*d*cos(d*x + c)^4 + 3*(a^8*b^2 + 3*a^7*b^3 + 3
*a^6*b^4 + a^5*b^5)*d*cos(d*x + c)^2 + (a^7*b^3 + 3*a^6*b^4 + 3*a^5*b^5 + a^4*b^6)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c)**2)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right )^{2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c)^2)^(7/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c)^2 + a)^(-7/2), x)