Optimal. Leaf size=391 \[ \frac{2 \left (-a^2 b^2 \left (45 c^2-49 d^2\right )+50 a^3 b c d-8 a^4 d^2-70 a b^3 c d+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^3 (b c-a d) \sqrt{a+b \tan (e+f x)}}-\frac{4 \left (-2 a^2 d+5 a b c+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac{i (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]
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Rubi [A] time = 2.09794, antiderivative size = 391, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {3567, 3649, 3616, 3615, 93, 208} \[ \frac{2 \left (-a^2 b^2 \left (45 c^2-49 d^2\right )+50 a^3 b c d-8 a^4 d^2-70 a b^3 c d+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^3 (b c-a d) \sqrt{a+b \tan (e+f x)}}-\frac{4 \left (-2 a^2 d+5 a b c+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac{i (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 3567
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{7/2}} \, dx &=-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 \int \frac{\frac{1}{2} \left (-6 b c d-a \left (5 c^2-d^2\right )\right )-\frac{5}{2} \left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)+2 d (b c-a d) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \, dx}{5 \left (a^2+b^2\right )}\\ &=-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{4 \int \frac{\frac{1}{4} (b c-a d) \left (15 a^2 c^2-15 b^2 c^2+40 a b c d-7 a^2 d^2+3 b^2 d^2\right )-\frac{15}{2} (b c-a d)^2 (a c+b d) \tan (e+f x)-d (b c-a d) \left (5 a b c-2 a^2 d+3 b^2 d\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx}{15 \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt{a+b \tan (e+f x)}}-\frac{8 \int \frac{-\frac{15}{8} (b c-a d)^2 \left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right )+\frac{15}{8} (b c-a d)^2 \left (3 a^2 b c^2-b^3 c^2-2 a^3 c d+6 a b^2 c d-3 a^2 b d^2+b^3 d^2\right ) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{15 \left (a^2+b^2\right )^3 (b c-a d)^2}\\ &=-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d)^2 \int \frac{1+i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac{(c+i d)^2 \int \frac{1-i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}\\ &=-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d)^2 \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b)^3 f}+\frac{(c+i d)^2 \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b)^3 f}\\ &=-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d)^2 \operatorname{Subst}\left (\int \frac{1}{i a+b-(i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(a-i b)^3 f}+\frac{(c+i d)^2 \operatorname{Subst}\left (\int \frac{1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^3 f}\\ &=-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a-i b)^{7/2} f}+\frac{i (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^{7/2} f}-\frac{2 (b c-a d) \sqrt{c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt{a+b \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 6.23535, size = 498, normalized size = 1.27 \[ -\frac{\frac{(c+d \tan (e+f x))^{3/2}}{(a-i b) (a+b \tan (e+f x))^{3/2}}-\frac{3 (c-i d) \left (\frac{\sqrt{c-i d} \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}-\frac{\sqrt{c+d \tan (e+f x)}}{(a-i b) \sqrt{a+b \tan (e+f x)}}\right )}{a-i b}}{3 f (b+i a)}+\frac{\frac{3 (c+i d) \left (\frac{\sqrt{-c-i d} \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2}}+\frac{\sqrt{c+d \tan (e+f x)}}{(a+i b) \sqrt{a+b \tan (e+f x)}}\right )}{a+i b}+\frac{(c+d \tan (e+f x))^{3/2}}{(a+i b) (a+b \tan (e+f x))^{3/2}}}{3 f (-b+i a)}+\frac{b (c+d \tan (e+f x))^{5/2}}{5 f (-b+i a) (b c-a d) (a+b \tan (e+f x))^{5/2}}-\frac{b (c+d \tan (e+f x))^{5/2}}{5 f (b+i a) (b c-a d) (a+b \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Fricas [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.
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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.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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